I want to prove that for $a,b,c>0$ we have

$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}= \frac a{\sqrt{a^2+3b^2+3c^2}}+\frac{b}{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}.$$

My first attempt: By Cauchy-Schwarz we have $$\left(\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}\right)^2\le3\sum_{cyc}\frac{a^2}{a^2+3b^2+3c^2}$$ so we only need to prove that the right-hand side is always less than $\frac{9}{7}$, but this is false. Failed

Second attempt: By Cauchy-Schwarz

$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=\sum_{cyc} \frac 1{\sqrt{1+3\frac{b^2}{a^2}+3\frac{c^2}{a^2}}}\le\sum_{cyc} \frac{\sqrt 7}{1+3\frac{b}{a}+3\frac{c}a}$$

so it remains to prove that $$\sum_{cyc} \frac{a}{1+3b+3c}\le\frac37$$ but this is wrong for example for $a=1,b=1,c=2$. Failed

Third attempt: Let $S=3(a^2+b^2+c^2)$. We need to prove $$\sum_{cyc} \frac{a}{\sqrt{S-2a^2}}\le \frac37.$$ But $x\mapsto \frac{x}{\sqrt{S -2x^2}}$ is convex so Jensen has the wrong direction...


Just an observation, the equality is achieved not only for $a=b=c$, but also for $$(a^2 \colon b^2 \colon c^2) = (8\colon 1\colon 1)$$


Define $$x=\frac{a^2}{a^2 + 3 b^2 + 3 c^2},\ y=\frac{b^2}{b^2 + 3 a^2 + 3 c^2},\ z=\frac{c^2}{c^2 + 3 a^2 + 3 b^2}$$

One checks (say by direct calculation) that $$28 x y z + 8(x y + x z + y z )+ x+y+z-1=0$$

So it is enough to show that the maximum of the function $\sqrt{x}+\sqrt{y}+\sqrt{z}$ on the part of the above surface in the first octant is $\frac{3}{\sqrt{7}}$. The Lagrange multiplier system $$28 x y z + 8(x y + x z + y z )+ x+y+z-1=0\\ t - x( 28 y z + 8 (y+z) + 1)^2 =0\\ t - y( 28 x z + 8 (x+z) + 1)^2 =0\\ t - z( 28 x y + 8 (x+y) + 1)^2 =0$$

is in fact not that hard to solve, if we use Groebner bases. First, by elimination we get the equation in $t$:

$$72313663744 t^7 - 207058475232 t^6 - 212349914280 t^5 + 806857109604 t^4 + 125825565483 t^3 - 784526490225 t^2=0$$ which factor nicely as $$t^2 (1372 t - 2025) (343 t - 729) (56 t + 81) (2744 t^2 - 1944 t - 6561)=0$$

Now one considers each of the possible positive values of $t$ and solves the system in $x$, $y$, $z$.

Case 1. $343 t - 729=0$. We get $x=y=z=\frac{1}{7}$

Case 2. $1372 t - 2025=0$. We get the solution $x=\frac{4}{7}$, $y=z=\frac{1}{28}$ and the cyclic permutation of it.

Case 3. $t = \frac{81(6 + 19\sqrt{2})}{1372}$

We get $x=\frac{-2 + 3 \sqrt{2}}{7}$, $y=z = \frac{-2 + 3 \sqrt{2}}{28}$ and the circular permutations.

Case 4. $t=0$ gives negative solutions, so we discard it

The inequality now follows, in the cases 1, 2, the functions takes the maximum value $\frac{3}{\sqrt{7}}$, in case 3 the value is smaller.


We need to prove that: $$\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\leq\frac{3}{\sqrt7},$$ where $a$, $b$ and $c$ are positive numbers.

Indeed, by C-S $$\left(\sum_{cyc}\sqrt{\frac{a}{a+3b+3c}}\right)^2\leq\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\sum_{cyc}(17a+2b+2c).$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{a}{(a+3b+3c)(17a+2b+2c)}\leq\frac{3}{49(a+b+c)}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that: $$\sum_{cyc}\frac{a}{(9u-2a)(6u+15a)}\leq\frac{1}{49u}$$ or $$49u\sum_{cyc}a(9u-2b)(9u-2c)(2u+5b)(2u+5c)\leq3\prod_{cyc}((9u-2a)(2u+5a)).$$ We'll prove that the last inequality is true even for any reals $a$, $b$ and $c$.

Indeed, since $\sum\limits_{cyc}a(9u-2b)(9u-2c)(2u+5b)(2u+5c)$ is a fifth degree polynomial,

the last inequality is equivalent to $f(w^3)\geq0,$ where $$f(w^3)=-3000w^6+A(u,v^2)w^3+B(u,v^2).$$ But $f$ is a concave function.

Thus, $f$ gets a minimal value for an extreme value of $w^3$, which happens for equality case of two variables.

Since the last inequality is an even degree, homogeneous and symmetrical, it's enough to assume $b=c=1,$ which gives $$\frac{a}{(a+6)(17a+4)}+\frac{2}{(3a+4)(2a+19)}\leq\frac{3}{49(a+2)}$$ or $$(a-1)^2(a-8)^2\geq0$$ and we are done!

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    $\begingroup$ Can I ask: when you use Cauchy-Schwarz, do you look for $x,y,z$ such that: $$\frac{7a}{(a+3b+3c)(xa+yb+zc)^2}=\frac{7b}{(b+3a+3c)(xb+ya+zc)^2}=\frac{7c}{(c+3a+3b)(xc+ya+zb)^2}$$ for both equality cases? $\endgroup$ – LHF Feb 23 '20 at 15:28
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    $\begingroup$ @Atticus Yes, of course. Because our vectors should be parallel. $\endgroup$ – Michael Rozenberg Feb 23 '20 at 15:33

Alternative proof:

Use the argument in https://artofproblemsolving.com/community/c6h1822770p12198977 or https://artofproblemsolving.com/community/c6h548438p3180154.

Let $x, y, z > 0$ such that $\frac{7a^2}{a^2 + 3b^2 + 3c^2} = x^2, \ \frac{7b^2}{b^2 + 3c^2 + 3a^2} = y^2, \ \frac{7c^2}{c^2 + 3a^2 + 3b^2} = z^2$. It is not hard to obtain $$F(x, y, z) = 4 x^2 y^2 z^2+8 x^2 y^2+8 x^2 z^2+8 y^2 z^2+7 x^2+7 y^2+7 z^2-49 = 0.$$

It suffices to prove that if $x, y, z > 0$ and $F(x,y,z) = 0$, then $x+y+z \le 3$. It suffices to prove that if $x, y, z > 0$ and $x + y + z > 3$, then $F(x, y, z) > 0$. Note that $F(\alpha x, \alpha y, \alpha z) > F(x, y, z)$ for any $\alpha > 1$ and $x, y, z > 0$. Thus, it suffices to prove that if $x, y, z > 0$ and $x+y+z = 3$, then $F(x, y, z) \ge 0$.

We use pqr method. Let $p = x + y + z = 3$, $q = xy + yz + zx$ and $r = xyz$. From $p^2 \ge 3q$, we have $q \le 3$. Let $q = 3(1-u^2)$ for $0 \le u\le 1$. We have \begin{align} (x-y)^2(y-z)^2(z-x)^2 &= -4p^3r+p^2q^2+18pqr-4q^3-27r^2\\ & = 108u^6 - 27(3u^2+r-1)^2 \end{align} which results in $108u^6 - 27(3u^2+r-1)^2 \ge 0$ and hence $r \le (2u+1)(1-u)^2 \le 3$. Thus, we have \begin{align} F(x, y, z) &= 7p^2-16pr+8q^2+4r^2-14q-49 \\ &= 4(6-r)^2+72u^4-102u^2-100\\ &\ge 4(6 - (2u+1)(1-u)^2)^2 + 72u^4-102u^2-100\\ &= 2u^2(2u^2-4u+9)(2u-1)^2\\ &\ge 0. \end{align} We are done.


We need to prove that:


Let's normalize with $a+b+c=3$. Then the inequality is equivalent with:

$$\sqrt{\frac{a}{9-2a}}+\sqrt{\frac{b}{9-2b}}+\sqrt{\frac{c}{9-2c}}\leq \frac{3}{\sqrt{2}}$$

Without loss of generality suppose that $a\le b\le c$. Then we have $a+b\leq 2$ and we will prove:

$$\sqrt{\frac{a}{9-2a}}+\sqrt{\frac{b}{9-2b}} \leq \sqrt{\frac{2(a+b)}{9-a-b}}$$

Squaring twice, this is equivalent with:

$$\frac{(a-b)^2[729+81(a+b)^2-486(a+b)-16ab(a+b)]}{(9-2a)^2(9-2b)^2(9-a-b)^2}\geq 0$$

We have $16ab(a+b)\leq 16(a+b)$ and

$$729+81x^2-502x\geq 0, \text{ when }x \leq 2$$

It remains to prove that:

$$\sqrt{\frac{14(3-c)}{6+c}}+\sqrt{\frac{7c}{9-2c}}\leq 3$$

Squaring twice this is equivalent with:

$$\frac{81(c-1)^2 (12 - 5 c)^2}{(9 - 2 c)^2 (6 + c)^2}\geq 0$$

which gives the two equality cases.


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