Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that$$\sqrt{5a+5b+8ab}+\sqrt{5c+5b+8cb}+\sqrt{5a+5c+8ac}\ge 3\sqrt{2}+2\sqrt{5}.$$ The equality case is $(0,1,1)$ but if we set the point $\left(\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3}\right),$ the $LHS-RHS \approx 0.$

I tried to square both side and obtain$$10(a+b+c)+8+2\sum_{cyc}\sqrt{5a+5b+8ab}\sqrt{5c+5b+8cb}\ge (3\sqrt{2}+2\sqrt{5})^2,$$but I have no clue to work with the yield $\sum_{cyc}\sqrt{5a+5b+8ab}\sqrt{5c+5b+8cb}.$

I also tried to use Holder inequality$$(LHS)^2.\sum_{cyc}(5a+5b+8ab),$$which did not help well.

Does mixing variables technique help here?

I hope we can find some brighter ideas. Thank you for your interest.

Updated edit.

We got some answers and progresses which seems not simple. The nice proof is teasing with us, isn't it ?


7 Answers 7


Let us fix first some notations:





$\mathcal{C}=\{(a,b,c)\in\mathbb{R}^3,0\leq a \leq 20,0\leq b \leq 20,0\leq c \leq 20\}$


For any triple $(a,b,c)\in\mathcal{D}$ which is not in $\mathcal{C}$, one of the real numbers $a$, $b$ and $c$ is greater than 20 and obviously we have:

$g\geq \sqrt{5a}+\sqrt{5b}+\sqrt{5c}>10>3\sqrt{2}+2\sqrt{5}$.

We are left to proving the inequality for $(a,b,c)\in\mathcal{C}_f$. Since $g$ is continuous and differentiable function of the variables $a$, $b$ and $c$ and since $\mathcal{C}_f$ is a compact subset of $\mathbb{R}^3$, there is a minimum for $g$ on $\mathcal{C}_f$ which is attained for at least one point.

If the point where minimum is achieved lies on the boundary, one of its coordinates is $0$ or $20$ (in which case g is at least 10). Let us suppose we have a coordinate equal to $0$, without loss of generality (by symmetry), we can assume $c=0$. When then need to have $ab=1$, so $a$ is not $0$ and $b=1/a$ which yields:

$g = \sqrt{5a}+\sqrt{5/a}+\sqrt{5a+5/a+8}$

It is easily determined that the minimum of this function for $a>0$ is met when $a=1$ and thus $g=3\sqrt{2}+2\sqrt{5}$.

We are left with points on the interior of $\mathcal{C}_f$. If the minimum is met in the interior, it has to be a local extremum. We will rely on the method of Lagrange Multiplier to search for extremal points. Such a point must verify that the gradients of $f$ and $g$ (as function of $a$, $b$ and $c$) are colinear. These gradients can be computed:


$\mathrm{Grad}(g)=(\frac{8b+5}{2r_1}+\frac{8c+5}{2r_2},\frac{8a+5}{2r_1}+\frac{8c+5}{2r_3},\frac{8a+5}{2r_2}+\frac{8b+5}{2r_3})$, where

$r_1 = \sqrt{5a+5b+8ab}$

$r_2 = \sqrt{5a+5c+8ac}$

$r_3 = \sqrt{5b+5c+8bc}$

The fact that both gradients are colinear translate into their cross product being null, which yields three equations (one per coordinate of the cross product). These equation involve square roots, so we apply the classical trick to multiply each one with its seven conjugates (i.e. applying all possible sign choices for the square roots). Last as noted by Claude Lebovici, one has $c=(1-ab)/(a+b)$ and we can substitute this value in the equations. All in all, we obtain three polynomial equations in $a$ and $b$ by taking the numerators. One can note that $a-b$ appears as a factor in one equation while $a^2+2ab-1$ appears in two others equations and corresponds to case where $b=c$. The other factors imply too many terms to be copied here but are provided by any algebra package on a modern computer.

For each pair of equations, using a resultant, one can eliminate the variable $b$ to get a polynomial in $a$. We can then take a polynomial gcd between the three univariate polynomial to conclude that $p(a)=0$ if $(a,b,c)$ is an extremum of $g$ on $\mathcal{C}_f$. When factoring over $\mathbb{Q}[a]$, the polynomial $p$ has 11 irreducible factors with degrees ranging from 1 to 5. We can then inject the factors (one after each other) in the three equations and a tedious case study (keeping only positive real numbers) identify the following local extrema:


$a=b$ are roots of the degree 5 five polynomial $512x^5+320x^4-300x^3-100x-125$ and $c=(1-ab)/(a+b)$

The extrema found are higher that the minimum previously identified. This concludes the identification of the potential minima of $g$ on $\mathcal{C}_f$. The inequality is thus proved since it corresponds to the minimum of $g$ on $\mathcal{C}$.

As a further note, this proof is very computational and not elegant. However, I believe the method can be applied to the other questions you submitted on this site.

  • $\begingroup$ +1, thanks. I will wait to accept the answer. The problem might be too hard for a simple proof. $\endgroup$
    – TATA box
    Commented Aug 4, 2023 at 15:30

A proof using Holder inequality.

By Holder inequality, we have \begin{align*} &\left(\sum_{\mathrm{cyc}} \sqrt{5a + 5b + 8ab}\right)^2 \sum_{\mathrm{cyc}} (5a + 5b + 8ab)^2\Big(5a + 5b + (6\sqrt{10} - 5)c\Big)^3\\ \ge{}& \left[\sum_{\mathrm{cyc}}(5a + 5b + 8ab)\Big(5a + 5b + (6\sqrt{10} - 5)c\Big)\right]^3. \tag{1} \end{align*}

It suffices to prove that \begin{align*} &\left[\sum_{\mathrm{cyc}} (5a + 5b + 8ab)\Big(5a + 5b + (6\sqrt{10} - 5)c\Big)\right]^3\\ \ge{}& \left(3\sqrt 2 + 2\sqrt 5\right)^2\sum_{\mathrm{cyc}} (5a + 5b + 8ab)^2\Big(5a + 5b + (6\sqrt{10} - 5)c\Big)^3.\tag{2} \end{align*}

We use the pqr method.

Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$. Using $p^2 \ge 3q = 3$, we have $p \ge \sqrt 3$.

(2) is written as $$(d_1 r + d_2)^3 \ge \left(3\sqrt 2 + 2\sqrt 5\right)^2(d_3 r^2 + d_4 r + d_5) $$ or \begin{align*} &d_1^3r^3 + \left(-12\sqrt{10}\, d_3 + 3d_1^2d_2 - 38d_3\right)r^2 + \left(-12\sqrt{10}\,d_4 + 3d_1d_2^2 - 38d_4\right)r\\[6pt] &\qquad - 12\sqrt{10}\,d_5 + d_2^3 - 38d_5 \ge 0 \tag{3} \end{align*} where \begin{align*} d_1 &= 144\,\sqrt {10}-240, \\[6pt] d_2 &= 50\,{p}^{2}+60\,\sqrt {10}+40\,p-100, \\[6pt] d_3 &= -92160\,p\sqrt {10}-950400\,\sqrt {10}+569600\,p+2832000, \\[6pt] d_4 &= 72000\,{p}^{3}\sqrt {10}-16000\,{p}^{4}+152550\,{p}^{2}\sqrt {10}- 150000\,{p}^{3}+28800\,p\sqrt {10}\\[6pt] &\qquad -449250\,{p}^{2}-495000\,\sqrt {10}+ 160000\,p+1475000, \\[6pt] d_5 &= 6250\,{p}^{5}+11250\,{p}^{3}\sqrt {10}+10000\,{p}^{4}-17000\,{p}^{3}+ 9000\,p\sqrt {10}+50000\,p. \end{align*}

We have $d_1^3 \ge 0$, and $-12\sqrt{10}\, d_3 + 3d_1^2d_2 - 38d_3 \ge 0$, and $-12\sqrt{10}\,d_4 + 3d_1d_2^2 - 38d_4 \ge 0$.

If $p^2 \ge 4q = 4$, we have $- 12\sqrt{10}\,d_5 + d_2^3 - 38d_5 \ge 0$. Thus, (3) is true.

If $p^2 < 4q = 4$, using degree three Schur $r \ge \frac{4pq - p^3}{9} = \frac{4p - p^3}{9}\ge 0$, we have \begin{align*} &d_1^3r^3 + \left(-12\sqrt{10}\, d_3 + 3d_1^2d_2 - 38d_3\right)r^2 + \left(-12\sqrt{10}\,d_4 + 3d_1d_2^2 - 38d_4\right)r\\[6pt] &\qquad - 12\sqrt{10}\,d_5 + d_2^3 - 38d_5\\[6pt] \ge{}&d_1^3\Big(\frac{4p - p^3}{9}\Big)^3 + \left(-12\sqrt{10}\, d_3 + 3d_1^2d_2 - 38d_3\right)\Big(\frac{4p - p^3}{9}\Big)^2\\[6pt] &\qquad + \left(-12\sqrt{10}\,d_4 + 3d_1d_2^2 - 38d_4\right)\Big(\frac{4p - p^3}{9}\Big)\\[6pt] &\qquad - 12\sqrt{10}\,d_5 + d_2^3 - 38d_5\\ \ge{}& 0. \end{align*} Thus, (3) is true.

We are done.

  • $\begingroup$ It turns out simpler. Thank you @RiverLi $\endgroup$
    – TATA box
    Commented Sep 16, 2023 at 8:54
  • $\begingroup$ @TATAbox Holder is powerful. Perhaps there is SOS solution for (2). By the way, as can be seen, here, LM or KKT conditions are not good to provide a solution which can be verified by hand. However, I think you or someone have tricky solution. $\endgroup$
    – River Li
    Commented Sep 16, 2023 at 9:33
  • 1
    $\begingroup$ I came up with the problem accidently but I have no original solution yet. It seems simple form. Hope to see more possible ideas in during contest time $\endgroup$
    – TATA box
    Commented Sep 16, 2023 at 11:07
  • $\begingroup$ I accepted your answer. It is the best proof, I think $\endgroup$
    – TATA box
    Commented Sep 17, 2023 at 13:54
  • $\begingroup$ @TATAbox Thanks. I tried $x = \sqrt{5a + 5b + 8ab}, \cdots$ and hope to find the relation among $x^2, y^2, z^2$ but failed. $\endgroup$
    – River Li
    Commented Sep 17, 2023 at 15:18

Using algebra

From the equality constraint $$c=\frac{1-a b}{a+b}$$ Now consider the function $$F(a,b)=\sqrt{\frac{a^2 (5-8 b)+8 a+5}{a+b}}+\sqrt{\frac{b^2(5-8 a) +8 b+5}{a+b}}+$$ $$\sqrt{8 a b+5 a+5 b}-\left(3 \sqrt{2}+2 \sqrt{5}\right) $$ and ask for a contour level of, say, $F(a,b)=0.01$ or smaller.

Now, compute $$\frac{\partial F(a,b)}{\partial a} \qquad \text{and} \qquad \frac{\partial F(a,b)}{\partial b}$$ (very nasty expressions) and use Newton-Raphson method starting with the same value $a_0=b_0=d$.

Varying $d$ from $0.01$ to $0.99$ by steps of $0.01$

  • for $0.01 \leq d \leq 0.70$, the solution is $a=b=\frac{1}{\sqrt{3}}$ which implies $c=\frac{1}{\sqrt{3}}$
  • for $0.71 \leq d \leq 0.99$, the solution is $a=b=0.8500134$ which would give $c=0.1632193$

Using the inverse symbolic calculator nothing was found for the numerical value of $a$ but the value of $c$ is found to be almost identical to the positive root of the quadratic $$760 x^2+8919 x-1476=0$$ (the absolute difference is $4.79\times 10^{-14}$). In other words $$a=b\sim\frac{8919-3 \sqrt{9337289}+\sqrt{165894562-53514 \sqrt{9337289}}}{1520}$$ $$c\sim\frac{3 \left(\sqrt{9337289}-2973\right)}{1520}$$ Back to $$G(a,b,c)=\sqrt{5a+5b+8ab}+\sqrt{5c+5b+8cb}+\sqrt{5a+5c+8ac}-\left(3 \sqrt{2}+2 \sqrt{5}\right)$$ the first set gives $$G(a,b,c)=\sqrt{6 \left(4+5 \sqrt{3}\right)}-3 \sqrt{2}-2 \sqrt{5}=8.14229\times 10^{-4}$$ and the second set (the too long expression is converted to decimals) gives $$G(a,b,c)=3.4495\times 10^{-2}$$


Another way to visualize the problem is to look at the contour plot of function $$H(a,b)=\left(\frac{\partial F(a,b)}{\partial a} \right)^2+\left(\frac{\partial F(a,b)}{\partial b} \right)^2$$

Assuming that the solutions of the partial derivatives correspond to $a=b$, after a few squaring, what is left, after tedious factorizations, is $$a^2 \left(3 a^2-1\right) \left(512 a^5+320 a^4-300 a^3-100 a-125\right)$$ $$ \left(1536 a^7+2240 a^6+188 a^5-320 a^4+400 a^3+125 a^2+100 a+125\right)=0$$

  • $a=0$ cannot be retained because of the definition of $c$
  • the pentic equation has three real roots $\{-1.12934,0.596038,0.850013\}$
  • the heptic equation has one real root $-0.745617$

For each root, we need to compute $$\sqrt{2} \left(\sqrt{-8 a^2+5 a+\frac{5}{a}+8}+\sqrt{a (4 a+5)}\right)-\left(3 \sqrt{2}+2 \sqrt{5}\right)$$ and exclude all the extra solutions introduced by the squaring steps.

Notice that $$-8 a^2+5 a+\frac{5}{a}+8 \geq 0$$ implies $$ 0 ~<~a ~\leq \frac{1}{24} \left(5+2 \sqrt{217} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{5885}{217 \sqrt{217}}\right)\right)\right)$$ which is $1.53880$.

  • $\begingroup$ (+1) Thank you for your interest. Why did you write "very nasty expressions" ? $\endgroup$
    – TATA box
    Commented Aug 2, 2023 at 22:47
  • $\begingroup$ @TATAbox. Write the partial derivatives to see ! Cheers :-) $\endgroup$ Commented Aug 3, 2023 at 4:02
  • $\begingroup$ what does that mean ? $\endgroup$
    – TATA box
    Commented Aug 3, 2023 at 4:17
  • 1
    $\begingroup$ @MichaelRozenberg. Even minus something $\endgroup$ Commented Aug 3, 2023 at 9:13
  • 3
    $\begingroup$ @TATA box In my opinion, it's not solution. $\endgroup$ Commented Aug 3, 2023 at 13:19


$$ f(a,b) = \sqrt{5(a+b)+8a b} $$


$$ F(a,b,c) = f(a,b) + f(a,c) + f(b,c) $$

for $a \ge 0, b \ge 0, c \ge 0$ we have

$$ \cases{ F(a,b,0)\le F(a,b,c)\\ F(0,b,c)\le F(a,b,c)\\ F(a,0,c)\le F(a,b,c) } $$

now searching for the minimum amongst

$$ \cases{ \min_{a,b} F(a,b,0) \ \ \ \text{s.t.}\ \ \ \{a \ge 0, b \ge 0, ab =1\}\\ \min_{a,c} F(a,0,c) \ \ \ \text{s.t.}\ \ \ \{a \ge 0, c \ge 0, ac =1\}\\ \min_{b,c} F(0,b,c) \ \ \ \text{s.t.}\ \ \ \{b \ge 0, c \ge 0, bc =1\}\\ } $$

we arrive at the same value $F(a^*,b^*,0) = F(a^*,0,b^*)= F(0,b^*,c^*) = 3 \sqrt{2}+2 \sqrt{5}$ for $a^*=b^*=c^* = 1$

  • $\begingroup$ Upvote for your try, but it is not full solution. $\endgroup$
    – TATA box
    Commented Aug 5, 2023 at 13:48

Some remarks for :


As @ClaudeLeibovici we have :


The function $f(x)$ is concave for $x\in[0,1]$ so we use a chord for $c\in[1,1.5]$ :

$$f(x)\ge g(x)=\frac{\left(f\left(2-c+\left(1-c\right)^{2}\right)-f\left(0\right)\right)}{2-c+\left(1-c\right)^{2}}\left(x-\left(2-c+\left(1-c\right)^{2}\right)\right)+f\left(2-c+\left(1-c\right)^{2}\right)$$

We can show that the function $g(x)$ is decreasing so we can set $x=2-c+\left(1-c\right)^{2}$ so now we have a one variable inequality we can use some bound .

For $c\in[0,1/\sqrt{3}]$ we have that around the minimum of $f(x)$ it's convex:

$$f(x)\geq h(x)=f'\left(c-\sqrt{3}\left(c-\frac{1}{\sqrt{3}}\right)\right)\left(x-c+\sqrt{3}\left(c-\frac{1}{\sqrt{3}}\right)\right)+f\left(c-\sqrt{3}\left(c-\frac{1}{\sqrt{3}}\right)\right)$$

Which show empirically that the minimum of $f(x)$ is greater than the proposed bound .

Also for $c\in[1/\sqrt{3},1],x\in[0,1]$ we have :

$$-\frac{3c}{\sqrt{3}}\left(g\left(0\right)-g\left(\frac{\sqrt{3}}{3c}\right)\right)\left(x-\frac{\sqrt{3}}{3c}\right)+g\left(\frac{\sqrt{3}}{3c}\right)-\sqrt{c+\left(\frac{1}{c}-1\right)x}\left(3\sqrt{2}+2\sqrt{5}\right)\geq 0$$

Where :


To be continued ...

  • $\begingroup$ Upvote for a nice try. Hope you full it soon $\endgroup$
    – TATA box
    Commented Aug 5, 2023 at 10:02
  • $\begingroup$ @TATAbox Thanks you I think we can do something here just with a line . $\endgroup$ Commented Aug 5, 2023 at 10:20
  • $\begingroup$ You are welcome to share it $\endgroup$
    – TATA box
    Commented Aug 5, 2023 at 10:32
  • $\begingroup$ Can you full it? I will accept and award it $\endgroup$
    – TATA box
    Commented Aug 6, 2023 at 10:22

See that, $$ \begin{align*} \left(\sqrt{5a+5b+8ab}+\sqrt{5c+5b+8cb}+\sqrt{5a+5c+8ac}\right)^2 \\ = 5a+5b+8ab+5c+5b+8cb+5a+5c+8ac + \\ 2\left (\sqrt{(5a+5b+8ab)(5c+5b+8cb)} \right) + 2\left (\sqrt{(5c+5b+8cb)(5a+5c+8ac)} \right) + 2\left (\sqrt{(5a+5b+8ab)(5a+5c+8ac)} \right) \\ = 10(a+b+c) + 8(ab+cb+ac) + 2\left (\sqrt{(5a+5b+8ab)(5c+5b+8cb)} \right) ...\\ \ge 10\sqrt{2}+8 + 2\left (\sqrt{(5a+5b+8ab)(5c+5b+8cb)} \right) ... \text{, as $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab + bc+ ca)... \ge 2$. Now,}\\\\ 2\left (\sqrt{(5a+5b+8ab)(5c+5b+8cb)} \right)\\ = 2\left (\sqrt{(25ac+25ab+40abc + 25bc+ 25b^2 + 40cb^2 + 40abc+40ab^2+64acb^2)} \right)\\ = 2\left (\sqrt{(25(ab+bc+ac) +40abc + 25b^2 + 40cb^2 + 40abc+40ab^2+64acb^2)} \right)\\ = 2\left (\sqrt{25 +40abc + 25b^2 + 40cb^2 + 40abc+40ab^2+64acb^2)} \right)\\ =2\left (\sqrt{25 +40abc + 25b^2 + 40b(cb + ac+ ab) +64acb^2)} \right)\\ =2\left (\sqrt{25 +40abc + 25b^2 + 40b +64acb^2)} \right)\\ \ge 2\sqrt{25} = 10 \\ \text{again because, a,b,c cannot all be 0, again because ab+bc+ca=1}, \\ \text{& Similar for other $2\sqrt{blobs}$, That gives that the square, }\\ \left(\sqrt{5a+5b+8ab}+\sqrt{5c+5b+8cb}+\sqrt{5a+5c+8ac}\right)^2 \\ \ge 10 + 10 + 10 + 8 + 10\sqrt{2} = 38 + 10\sqrt{2},\\\text{That's the best I got..:-) + for you to continue..to prove $\ge 38 + 12\sqrt{10}$ result + } \\ \text{it's only possible positive root is the answer...} \end{align*}$$

  • $\begingroup$ There is something wrong. $\endgroup$
    – TATA box
    Commented Aug 5, 2023 at 4:22
  • $\begingroup$ Upvote for your try, but it is not full solution. $\endgroup$
    – TATA box
    Commented Aug 5, 2023 at 13:48

This answer is trying to answer the question with a full solution, but moreover the questions behind the question, which are:

  • $\bbox[yellow]{(0)}$ General discussion, which is the right framework?
  • $\bbox[yellow]{(1)}$ Which is the minimal value of the function $f:[0,\infty)^3\to\Bbb R$, $$ f(a,b,c)=\sum_{\text{cyclic}}\sqrt{5a+5b+8ab}\ , $$ constrained to the relation / on the subset of all $(a,b,c)$ with $$ 1 = ab+bc+ca\ ? $$
  • $\bbox[yellow]{(2)}$ Which is the complexity involved in the computation?
  • $\bbox[yellow]{(3)}$ Which methods are available to investigate humanly this minimal value?

$\bbox[yellow]{(0)}$ In a first part i will use CAS (computer algebra software, sage) in order to have a first firm answer with intermediate check points that may serve as a further orientation. The complexity will be transparent from the computer data. It remains to draw conclusions for the human point of view, and go into the one or other way to humanly attack the problem.

First of all, for the given kind of problem we have already a method, it uses Lagrange multiplicators, and universities do want its applications (unfortunately only in cheaper situations), and sites like MSE insist to avoid it (arguably because it is not elementary enough, pretending that an olimpiad-like solution is only wanted. Yes, if there is such a solution it is welcome. But notice that very often a problem like ours here will never survive into the final list of an NMO / IMO, say. Except for special counties, i've got such problems in the live stream. And there are reasons for this elimination.)

In my eyes, there is no reason for rejecting such an efficient analysis tool in problems of extremum in situations like ours here.

So let us combine the human and the computer force to have a first verdict.

$\bbox[yellow]{(1)}$ We associate the function $F:(0,\infty)^3\times\Bbb R$ defined by $$ F(a,b,c,m)=f(a,b,c) - m(ab+bc+ca-1)\ . $$ Its $\inf$ for points on the domain $$ D=\{\ (a,b,c)\ :\ a,b,c>0\ ,\ ab+bc+ca=1\ \} $$ are

  • either at the boundary of $D$, i.e. points with $abc=0$, $ab+bc+ca=1$,
  • or points of $(0,\infty)^3\times \Bbb R$ where $F'=(F'_a,F'_b,F'_c,F'_m)$ vanishes. (The condition $F'_m=0$ makes such a point belong to $D$.)

$\bbox[lightyellow]{(1a)}$ Let us examine the boundary points first. Using the symmetry, we may and do assume $c=0$. The constraint on $(a,b,c)$ implies $ab=1$. Then $f(a,b,c)/\sqrt 5$ is $$ \sqrt{a+b+sab} + \sqrt{a}+\sqrt{b}\ , $$ with $\bbox[lightgreen]{\ s=8/5\ }$ for short, and we minimize this function. Apply again Lagrange multipliers.Let $G$ be the function $$(a,b,m)\to F(a,b,0,m)/\sqrt 5 =G(a,b,m) =\sqrt{a+b+s} + \sqrt {a}+\sqrt {b}-m(ab-1)\ , $$ and work with the condition $ab=1$. Boundary (with $ab=1$) examination. For $a\to 0$ or $b\to 0$ we have $b\to\infty$ respectively $a\to\infty$, so $F(a,b,0,m)\to\infty$, no infimum at the boundary. For the local extremal points we ask for $G'_a=G'_b=G'_m=0$. The condition $G'_m=0$ is $ab=1$, i.e. $a=1/b$. We obtain: $$ \begin{aligned} \frac 1{2\sqrt{a+b+s}} +\frac 1{2\sqrt a}-mb &=0 =\frac 1{2\sqrt{a+b+s}} +\frac 1{2\sqrt b}-ma \ ,\text{ i.e.} \\ \frac a{\sqrt{a+b+s}} + \sqrt a &=2m =\frac b{\sqrt{a+b+s}} +\sqrt b\ . \end{aligned} $$ We ignore $m$, use the equation resulted from above by subtracting the right part from the left side, and can factor $\sqrt a-\sqrt b$. The remained factor is $\frac {\sqrt a+\sqrt b}{\sqrt{a+b+s}} + 1>0$, so there is only one point that comes in discussion. It must correspond to the one absolute minimum (which exists, since the boundary discussion shows that $G\to \infty$ at boundary). So $f(a,b,0)$ is minimal for $a=b=1$, we obtain the value $3\sqrt 2+2\sqrt 5$ for this point. This is a proof for the minimum value on the boundary so far.

$\bbox[lightyellow]{(1b)}$ We will compare this value with local extremal (minimal) values. When computing derivatives it is useful to use $f(a,b,c)/\sqrt 8$, which is the cyclic sum of the expression $\sqrt{ab+s(a+b)}$, where this time $\bbox[lightgreen]{\ s=5/8\ }$ for short. Then $f$ and $f/\sqrt 8$ are minimal at the same points. We use the latter version, for typing reasons. Instead of $F$ we have its version
We set and compute: $$ \begin{aligned} F(a,b,c;m)&:= \underbrace{\sqrt{bc+s(b+c)}}_{\text{Notation: }x=x(a,b,c)} + \underbrace{\sqrt{ca+s(c+a)}}_{\text{Notation: }y=y(a,b,c)} + \underbrace{\sqrt{ab+s(a+b)}}_{\text{Notation: }z=z(a,b,c)} -m(bc+ca+ab-1)\ ,\\ &\qquad\text{ Then the system $0=F'_a=F'_b=F'_c=F'_m$ has the equations:} \\ 0&=\frac 1{2x}\cdot 0 +\frac 1{2y}(c+s) +\frac 1{2z}(b+s) - m(b+c)\ ,\\ 0&=\frac 1{2x}(c+s) +\frac 1{2y}\cdot 0 +\frac 1{2z}(a+s) - m(c+a)\ ,\\ 0&=\frac 1{2x}(b+s) +\frac 1{2y}(a+s) +\frac 1{2z}\cdot 0 - m(a+b)\ ,\\ 1 &=bc+ca+ab\ . \end{aligned} $$ The first three equations may be seen as a linear system in $1/(2x)$, $1/(2y)$, $1/(2z)$, $$ \begin{bmatrix} 0 & c+s & b+s\\ c+s & 0 & a+s\\ b+s & a+s & 0 \end{bmatrix} \begin{bmatrix} 1/(2x)\\ 1/(2y)\\ 1/(2z) \end{bmatrix} = m \begin{bmatrix} b+c\\ c+a\\ a+b \end{bmatrix} \ , $$ the involved $3\times 3$-matrix has determinant $2(a+s)(b+s)(c+s)$, and its solution isolates the corresponding values with this common denominator, so we obtain four equations in $a,b,c;m$ of the shape: $$ \begin{aligned} \frac1{2\sqrt{bc+s(b+c)}} =\frac 1{2x} &=\frac{m\color{gray}{(a+s)}(bc+as)}{\color{gray}{(a+s)}(b+s)(c+s)}\ , \\ \frac1{2\sqrt{ca+s(c+a)}} =\frac 1{2y} &=\frac{m\color{gray}{(b+s)}(ca+bs)}{(a+s)\color{gray}{(b+s)}(c+s)}\ , \\ \frac1{2\sqrt{ab+s(a+b)}} =\frac 1{2z} &=\frac{m\color{gray}{(c+s)}(ab+cs)}{(a+s)(b+s)\color{gray}{(c+s)}}\ ,\\ 1 &= bc+ca+ab\ . \end{aligned} $$ We can eliminate $m$ now, there are three expressions equal to the square of $(a+s)(b+s)(c+s)/(2m)$, well, squaring may introduce solutions (corresponding to $\pm m$, but humanly we do so), and obtain: $$ (bc+s(b+c))(a+s)^2(bc+as)^2 = (ca+s(c+a))(b+s)^2(ca+bs)^2 = (ab+s(a+b))(c+s)^2(ab+cs)^2 \ , $$ this leads to two equations of degree six in $a,b,c$.

Subtracting any two of the three parts of the double equality introduces one of the factors $(a-b)$, $(b-c)$, $(c-a)$, but the remained factor has huge degree for the human hand.

It is time to face the reality, let us ask the machine which are the real positive solutions of the system. This is a giant leap for the man, but a small one for the mankind, so let us keep it small:

R.<a,b,c,X,Y,Z,m> = PolynomialRing(QQ)
# we are using X, Y, Z instead of 1/x, 1/y, 1/z
S.<t> = PolynomialRing(QQ)

s = 5/8
J = R.ideal([
        X/2*0       + Y/2*(c + s) + Z/2*(b + s) - m*(b + c) ,
        X/2*(c + s) + Y/2*0       + Z/2*(a + s) - m*(c + a) ,
        X/2*(b + s) + Y/2*(a + s) + Z/2*0       - m*(a + b) ,
        -1 + X^2*(b*c + s*(b + c)) ,
        -1 + Y^2*(c*a + s*(c + a)) ,
        -1 + Z^2*(a*b + s*(a + b)) ,
        -1   + b*c + c*a + a*b     , ])

pa = J.elimination_ideal([X,Y,Z, b,c, m]).gens()[0]
print(f'After elimination of all other variables, a satisfies the equation:')
print(f'0 = {pa.subs(a=t).factor()}')

And the prints deliver the complexity of the problem:

After elimination of all other variables, a satisfies the equation:
0 = (3221225472000) 
    * (t^2 - 1/3)
    * (t^3 - 21/8*t^2 + t + 45/32)
    * (t^3 + 11/8*t^2 + t - 5/32)
    * (t^5 + 7/80*t^4 + 501/512*t^3 + 421/1280*t^2 - 4021/25600*t + 25481/2048000)
    * (t^5 + 5/8*t^4 - 75/128*t^3 - 25/128*t - 125/512)

And each of the involved non-constant factor has a positive root in $(0,\infty)$, we have expected the root $a=1/\sqrt 3$, from $t^2-1/3=0$, but also other factors come with positive roots:

sage: for f, mul in pa.subs(a=t).factor():
....:     print(f'{f}\n\tvanishes in:\n\t{[r for r in f.roots(ring=AA, multiplicities=False) if 0 < r < 1]}')
t^2 - 1/3
    vanishes in:
t^3 - 21/8*t^2 + t + 45/32
    vanishes in:
t^3 + 11/8*t^2 + t - 5/32
    vanishes in:
t^5 + 7/80*t^4 + 501/512*t^3 + 421/1280*t^2 - 4021/25600*t + 25481/2048000
    vanishes in:
    [0.12193648467888863?, 0.1632193275966720?]
t^5 + 5/8*t^4 - 75/128*t^3 - 25/128*t - 125/512
    vanishes in:

A "manual" consideration has to overcome with thus difficulties.

$\bbox[yellow]{(2)}$ The involved complexity is thus too much for the pen and paper approach. Even with a clever handling of the above system in $a,b,c;m$, we still have to consider (algebraically) complicated points. One speculation would be that some positivity argument may rule out some of the many possible $a$-values above. Well, let us see which are all the solutions of the given system. We are again using computer support, alternatively, we use the system in $a,b,c$ that we have finally obtained, and restrict to positive solutions.

For each solution we also compute the value of $f$ in it. Code:

s = 5/8
R.<a,b,c> = PolynomialRing(QQ)

def E(a, b, c):    
    return (b*c + s*(b + c)) * (a + s)^2 * (b*c + a*s)^2

J = R.ideal([
        a*b + b*c + c*a - 1
        , E(a,b,c) - E(b,c,a), E(a,b,c) - E(c,a,b)])
for point in J.variety(ring=AA):
    A, B, C = point.values()
    val = ( sqrt(8*A*B + 5*A + 5*B) +
            sqrt(8*B*C + 5*B + 5*C) +
            sqrt(8*C*A + 5*C + 5*A) )
    if A >= B and B >= C and C > 0: 
        print("The following point (a, b, c) is an ordered solution:")
        print(f"\ta = {A} with minimal polynomial = {A.minpoly()}")
        print(f"\tb = {B} with minimal polynomial = {B.minpoly()}")
        print(f"\tc = {C} with minimal polynomial = {C.minpoly()}")
        print("The value of the initial function in this point is:")
        print(f"\tf(a, b, c) = {val} with minimal polynomial {val.minpoly()}\n")

This gives:

The following point (a, b, c) is an ordered solution:
    a = 0.5773502691896258? with minimal polynomial = x^2 - 1/3
    b = 0.5773502691896258? with minimal polynomial = x^2 - 1/3
    c = 0.5773502691896258? with minimal polynomial = x^2 - 1/3
The value of the initial function in this point is:
    f(a, b, c) = 8.71559087079392? with minimal polynomial x^4 - 48*x^2 - 2124

The following point (a, b, c) is an ordered solution:
    a = 0.8500133943883169? with minimal polynomial = x^5 + 5/8*x^4 - 75/128*x^3 - 25/128*x - 125/512
    b = 0.8500133943883169? with minimal polynomial = x^5 + 5/8*x^4 - 75/128*x^3 - 25/128*x - 125/512
    c = 0.1632193275966720? with minimal polynomial = x^5 + 7/80*x^4 + 501/512*x^3 + 421/1280*x^2 - 4021/25600*x + 25481/2048000
The value of the initial function in this point is:
    f(a, b, c) = 8.74927159655177? with minimal polynomial x^10 - 163/4*x^8 - 568387/256*x^6 - 613979/32*x^4 - 6078993/4*x^2 - 4578338

Both values (possible local extremal values) are greater than the boundary minimum $f(0,1,1)=f(1,0,1)=f(1,1,0)$:

sage: QQbar(sqrt(18) + 2*sqrt(5))

So the absolute minimum is the boundary minimum $3\sqrt 2+2\sqrt 5$.

$\bbox[yellow]{(3)}$ Can we do better with human eyes and hands? One idea would be to use convexity arguments. But unfortunately near the boundary, especially near the minimal point, we have in all directions an increasing function, and "somewhere in the middle" in $(m,m,m)$ with $m=1/\sqrt3$ we also have a local minimum. So there will be portions (on the line / on any path) from $(0,1,1)$ to $(m,m,m)$ of concavity. It is hard to make them explicit.

Maybe a picture of the situation may help to visualize the problem.

mse problem 4743736

mse problem 4743736

This is one and the same spherical plot picture shown from different angles. Our function $f$ is red, but to make the picture focus on the critical part the plot was done for $\max(10, f)$, this red plot has full opacity. A spherical plot of a (positive) constant function is a part of the sphere with the radius given by the constant. For orientation we have in the same pictures such spherical constant plots for the following values:

  • ten, the gray truncation of $f$,
  • 8.714776642118, which corresponds numerically to (almost) $\min f$ taken at the boundary, it is green, and we allow some transparency to see through it,
  • and finally 8.716, the cyan piece of the sphere, which is opaque.

The intransparent red and cyan surfaces show where we expect the locally minimal value. In that portion of the plot, red, cyan, and green are almost indistinguishable as surfaces, we need colors to observe the difference.

The picture was computed as follows.

def E(a, b): return sqrt(8*a*b + 5*a + 5*b)
def f(a, b, c): return E(b, c) + E(c, a) + E(a, b)

def myplot(u, v):
    A, B, C = cos(u)*sin(v), sin(u)*sin(v), cos(v)
    den = sqrt(B*C + C*A + A*B)
    a, b, c = A/den, B/den, C/den    # thus bc + ca + ab = 1
    return min(10, f(a, b, c))

def   ten(u, v):    return 10
def mymin(u, v):    return 8.714776642118
def mymid(u, v):    return 8.716

eps = 0.000000000000001
R = (0 + eps, pi/2 - eps)
g1 = spherical_plot3d(myplot, R, R, color='red', opacity=1)
g2 = spherical_plot3d(   ten, R, R, color='lightblue', opacity=0.2)
g3 = spherical_plot3d( mymin, R, R, color='green', opacity=0.5)
g4 = spherical_plot3d( mymid, R, R, color='cyan', opacity=1)
(g1 + g2 + g3 + g4).show(aspect_ratio=1)

The above lines were presenting a full computer aided solution to the problem. Is there any human way to go? I have to doubt convexity arguments, since there cannot be any convexity - at least near the absolute minimal values at the boundary. Also, going algebraically is also a hard elimination task. Critical points of algebraic degree five (as algebraic numbers, numbers in $\bar{\Bbb Q}$) do occur, and we have to say something about these numbers, yes, the true local minimum is "a" simple algebraic number $1/\sqrt3$ (repeated three times on the components), but we still have to compute it and compare its value with the other critical points of degree five, or at least compute the Hessian / the convexity in these degree five critical points. So there is no "easy" way to go for a human (that fits the reasonable framework), i am for legitimating computer assistance in such cases. (Well, i would have done it first also without legitimation.)

  • $\begingroup$ Oh my god! I've never seen something complicated like that before $\endgroup$
    – TATA box
    Commented Sep 17, 2023 at 13:53
  • $\begingroup$ Which is that complicated point? @TATAbox $\endgroup$
    – dan_fulea
    Commented Sep 17, 2023 at 20:56
  • $\begingroup$ (The most complicated one, if there are a lot of them...) All the above can be made short. The solution is written not only for the OP, but also for the community. I am explaining the method used, Lagrange multiplicators, and the way to make this work. This involves going into equations of higher degree. In order to solve the system equations we eliminate, easy to say, hard to do humanly. After elimination of two variables we get one equation in one variable. Do you believe, when somebody claims which are the factors of the one polynomial? @TATAbox $\endgroup$
    – dan_fulea
    Commented Sep 17, 2023 at 21:05
  • 1
    $\begingroup$ Simply, my knowledge is not good enough to full understand your answer. It seems interesting, @danfulea $\endgroup$
    – TATA box
    Commented Sep 18, 2023 at 1:11

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