Let $x+y+z=3,x,y,z\ge 0$,show that $$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$

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I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that $$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge 6.$$

Without loss of generality we can let $x=\max{\{x,y,z\}}$

Proof: case 1 $x\ge y\ge z$

we can easily prove $$\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge\sqrt{y^2+x+2}+\sqrt{z^2+z+2}$$

and $$\sqrt{x^2+y+2}+\sqrt{y^2+x+2}\ge\sqrt{x^2+x+2}+\sqrt{y^2+y+2}$$

so we have $$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge \sqrt{x^2+x+2}+\sqrt{y^2+y+2}+\sqrt{z^2+z+2}.$$

Then use $$\sqrt{x^2+x+2}\ge\dfrac{3}{4}x+\dfrac{5}{4}$$ $$\sqrt{y^2+y+2}\ge\dfrac{3}{4}y+\dfrac{5}{4}$$ $$\sqrt{z^2+z+2}\ge\dfrac{3}{4}z+\dfrac{5}{4}$$ to get the result. Whereas the case 2 when $x\ge z\ge y$ can be proved using the same methods.

Now,I have another idea: using Holder inequality we have $$\left(\sum\sqrt{x^2+yz+2}\right)^2\left(\sum\dfrac{x^2+2yz+9}{x^2+yz+2}\right)\ge 36^3$$ $$\Longleftrightarrow \sum\dfrac{x^2+2yz+9}{x^2+yz+2}\le 1296$$

and the following link has some discussion about this problem http://www.artofproblemsolving.com/Forum/viewtopic.php?t=538230

and http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=538752&p=3097872#p3097872

and Vasc gave the hint:

$$\sum\sqrt{8(a^2+bc+2)}\ge \sum\sqrt{(3a+b+c)^2+7}\ge 12\sqrt 2$$

How prove this hint?Thank you everyone.

and my other idea is as follows:

let $a=\min(a,b,c)$ we can prove $$\sqrt{b^2+ca+2}+\sqrt{c^2+ab+2}\geq \sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\tag{1}$$ \begin{align*} &\sqrt{a^2+bc+2}+\sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\\ \geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8} \tag{2} \end{align*} Summing up \begin{align*} &\sum_{cyc}{\sqrt{a^2+bc+2}}\\ \geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8}\\ =&\sqrt{a^2+\frac{(3-a)^2}{4}+2}+\sqrt{(3-a)^2+2a(3-a)+8} \end{align*}

By the way: someone said $(1)$ is wrong? why? can anyone give an example? And hopefully someone can use this method to prove this inequality? Thank you very much!

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    $\begingroup$ I edited some of the English...for everyone to better understand. $\endgroup$ – Shuhao Cao Jun 13 '13 at 18:34
  • $\begingroup$ @math110 : Can you give a lower bound for $xy+yz+zx$? I think it will solve the problem. All I can think of is to use Lagrange's multiplier method, but can't find an elementary method to get a lower bound. $\endgroup$ – yojusmath Jun 16 '13 at 7:18
  • $\begingroup$ Thank you, You meaning $LHs\ge 2(xy+yz+xz)?$ can you share your solution ? $\endgroup$ – math110 Jun 16 '13 at 8:39
  • $\begingroup$ @mathmansujo The problem is that you can't give a lower bound for $xy+yz+zx$. Choose $x=y=0$, $z=3$. $\endgroup$ – Beni Bogosel Jun 20 '13 at 9:07
  • $\begingroup$ Not a solution. but I found it interesting that the upper bound can also be found as $\frac{3\sqrt{17}}2.$ $\endgroup$ – karakfa Jun 20 '13 at 15:45

Here is an explanation of Vasc's hint that you mentioned, which essentially solves the problem. Reproducing it here as: $$\sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq \sum _{\text{cyc}} \sqrt{(3x+y+z)^2+7} = \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}\geq 12\sqrt{2}$$

Now the last inequality follows immediately from Minkowski's inequality and $\sum x=3$, so I will focus only on the first inequality below.

The inequality $\displaystyle \sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}$ can be established through Schur-concavity of the vector function $f{(\color{blue}{u}}) = \sqrt{u_1}+\sqrt{u_2}+\sqrt{u_3}$. This function is symmetric and concave, hence Schur-concave. Alternately it is not difficult to establish the condition $$\left(u_i- u_j\right)\left(\partial _{u_i}f-\partial _{u_j}f\right) = -\frac{\left(u_i- u_j\right){}^2}{\sqrt{u_i u_j}} \leq 0.$$

<< If you are not familiar with Schur-concave property, it should be possible to derive similar results using the lemma $\sqrt{x-\epsilon }+\sqrt{y+\epsilon }\geq \sqrt{x}+\sqrt{y}$. This is easily shown to hold if $x-y \ge \epsilon \ge 0$ by squaring. Using the sequence $$[x, y, z] \succ [x - u, y + u, z] \succ [x - u, y + u - (v + u), z + (u + v)]$$ and applying the lemma twice, one can establish $$\sqrt{x-u}+\sqrt{y-v}+\sqrt{z+u+v}\geq \sqrt{x}+\sqrt{y}+\sqrt{z}$$ and the conditions needed.>>

As $f{(\color{blue}{u}})$ is Schur-concave, if we have $\color{blue}{a} \text{ and } \color{blue}{b}$ two vectors such that $\color{blue}{a}$ majorizes $\color{blue}{b}$, i.e. $\color{blue}{a} \succ \color{blue}{b} \implies f(\color{blue}{a})\leq f(\color{blue}{b})$. In this case we need to show that considered as vectors, $$\left[(2x+3)^2+7,(2y+3)^2+7, (2z+3)^2+7\right] \succ \left[8\left(x^2+\text{yz}+2\right),8\left(y^2+\text{zx}+2\right), 8\left(z^2+\text{xy}+2\right)\right]$$

If $s = xy+yz+zx$, we have $$\sum _{\text{cyc}} \sqrt{8\left(x^2+\text{yz}+2\right)}= \sum _{\text{cyc}} \sqrt{8x^2+8\text{yz} + (x+y+z)^2+7} = \sum _{\text{cyc}} \sqrt{(3x+y+z)^2+7+ 4(2\text{yz}- \text{xy}-\text{zx})} = \sum _{\text{cyc}} \sqrt{(2x+3)^2+7+ 4(3\text{yz}-s)}$$

Let $\displaystyle a = (2x+3)^2+7, b = (2y+3)^2 + 7 \text{ and } c = (2z+3)^2+7$. Also let $u = 4(s-3\text{yz}) \text{ and } v = 4(s-3\text{zx})$. Now $\displaystyle x \geq y \geq z \Longrightarrow \text{xy} \geq \text{zx} \geq \text{yz} \Longrightarrow u \geq 0$ and $u+v \geq 0$ and the deviations are in reverse order to the ordering of $a, b, c$- hence giving the conditions needed for majorization. It may be noted that in a crude sense, the deviations have mean $0$, and are ordered so as to bring the components "closer", hence increasing the concave function's value.

As $[a, b, c] \succ [a-u, b-v, c+(u+v)]$, we have $\displaystyle \sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}$.

Hope there is a simpler method to establish this. Another possible approach could be to directly check if $\displaystyle F(x, y, z) = \sum _{\text{cyc}} \sqrt{x^2+ \text{yz} + 2}$ is Schur-convex, in which case any allowable $[x, y, z]\succ [1, 1, 1]$ and the entire proof is done. $F$ is symmetric, but we also need to check if $x-y)\left(\partial _xF -\partial _yF\right)\geq 0$ for this.


Let's solve

$$ \min \Bigg[\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2} \Bigg]$$ subject to

$$ x+y+z=3 \\ x\ge0 \\ y\ge 0 \\ z\ge 0 $$

Formally, this is done with Kuhn-Tucker condition. Write those down, exploit symmetries, note that non-negativity constraints do not bind.

Solution: $x=y=z=1$. Hence, the minimum value of the objective is $6$.

The other inequality admits the same proof.

details: see wiki or original paper for details of KKT conditions. $$ L(x,y,z,\lambda,\mu_1,\mu_2,\mu_3) = -\Bigg[\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2} \Bigg] + \lambda[x+y+z=3] + \mu_1[x-0] +\mu_2[y-0] +\mu_3[z-0] $$ KKT conditions: \begin{align} [x]\qquad -\frac{1}{2}\eta_1 2x -\frac{1}{2}\eta_2 z-\frac{1}{2}\eta_3 y &= \lambda - \mu_1 \\ [y]\qquad -\frac{1}{2}\eta_1 z -\frac{1}{2}\eta_2 2y-\frac{1}{2}\eta_3 x &= \lambda - \mu_2 \\ [z]\qquad -\frac{1}{2}\eta_1 y -\frac{1}{2}\eta_2 x-\frac{1}{2}\eta_3 2z &= \lambda - \mu_3 \\ [\lambda]\qquad\qquad\qquad\qquad\quad x+y+z&=3\\ [\mu_1]\qquad \mu_1\ge 0\qquad\text{and}\qquad \mu_1x&=0 \\ [\mu_2]\qquad \mu_2\ge 0\qquad\text{and}\qquad \mu_2y&=0 \\ [\mu_3]\qquad \mu_3\ge 0\qquad\text{and}\qquad \mu_3z&=0 \\ \end{align} where $\eta_1 = (x^2+yz+2)^{-1/2},\eta_2 = (y^2+zx+2)^{-1/2}$ and $\eta_1 = (z^2+xy+2)^{-1/2}$

KKT methods states that the minimum of the objective should be a solution (for some $\lambda,\mu_1,\mu_2,\mu_3$) to the above system.

Wlog, let $x=\max\{x,y,z\}$ so that $x>0$ and $\mu_1=0$.

Case 1: $y=z=0$. Solve for $\lambda$ from $[x]$ and note that $\mu_2 <0$, therefore this case is empty.

Case 2: $y=0$ and $z=3-x\ne0$. Then $\mu_3=0$. Solve for $\lambda$ in $[x],[z]$, equate the two expressions and deduce that $x=z$. Observe that $\mu_2<0$ (at $x=z$), therefore this case is empty.

Case 3: $\mu_1=\mu_2=\mu_3=0$ so that non-negativity constraints are slack. Equate LHS of conditions $[x],[y],[z]$, combine with condition $[\lambda]$ to solve for $x,y,z$.

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    $\begingroup$ math110, I believe, I gave a formal proof to the problem you posed. I also believe, that this is a good approach to proving inequalities of the the form $f(x,y,z)\ge a$ subject to $g_i(x,y,z)\ge 0$ and $h_i(x,y,z)=0$. It's sharper, because CS holds with equality only under linearity. And thanks to downgrading my answer too. $\endgroup$ – mStudent Jun 15 '13 at 15:33
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    $\begingroup$ Thank you,But can you delete your anwser? becasuse some one see have one anwser,There Out of thinking about the problem of motivation.Thank you $\endgroup$ – math110 Jun 17 '13 at 7:58
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    $\begingroup$ @math110: It is a good answer. Why delete it? If there is another method, it will be posted in another answer, but you cannot request that all answers that are not as you want should be deleted, just so that your question will be listed on the unanswered list... $\endgroup$ – Beni Bogosel Jun 20 '13 at 9:03
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    $\begingroup$ @mStudent: How do you get all those $\mu_2<0$? In fact, I believe this is not true. The function's behavior is such that it attains its global minimum outside of the positive box, and, therefore, you have to MANUALLY check different minima (constrainted and unconstrainted). For example, if you set x=3.5,y=-0.5,z=0.0, then the objective value is LESS than 6. Hence, along the boundaries the function tends to its global minimum outside of the region. And $\mu_2$ (which is "tendency" of the objective function) can definitely be positive when $z=0$. Moreover, it is even when $y=z=0$. $\endgroup$ – Vadim Jun 20 '13 at 19:00
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    $\begingroup$ This is what typically makes KT huge. It is a universal method, but in practice it usually leads to lots of different cases with large systems of non-linear equations in each case. Even in the problem above to find just one internal solution (which basically becomes Lagrangian: $\mu_i=0$) one has to solve a system of nonlinear equations, which is not so easy to do. It is great that this universal method was mentioned for this problem, but the solution makes it seem to be much simpler than it is. $\endgroup$ – Vadim Jun 20 '13 at 19:17

Here is my sketch, mostly to convince that alternative approach (i.e. inequalities instead of search for minimum) really may be possible. At the other hand, useful inequalities have "differential" nature (that is, I think they may be most easily proven by differentiation, but there may be other ways to proof).

Let us denote $a=x^2+yz, b=y^2+zx, c=z^2+xy$; then

$$a+b+c=x^2+yz+y^2+zx+z^2+xy=x^2+y^2+z^2+\frac{1}{2}((x+y+z)^2-(x^2+y^2+z^2))=$$ $$=x^2+y^2+z^2+4.5-\frac{1}{2}(x^2+y^2+z^2)\ge\frac{1}{2}3+4.5=6$$

Starting from reasonably looking inequality $\sqrt{u+2}+\sqrt{v+2}+\sqrt{w+2}\ge\sqrt{u+v+w+18}$ (where $u,v,w\ge 0$) we let $u=a,v=b,w=c$ and conclude that $$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge\sqrt{24}=\frac{6}{\sqrt{3/2}}$$

Starting from another inequality $f(u,v,w)\equiv\sqrt{u+2}+\sqrt{v+2}+\sqrt{w+2}\ge g(u,v,w)\equiv\sqrt{3(u+v+w)+18} $ (where $u,v,w\ge 0$ and $u\approx v\approx w$) we let $u=a,v=b,w=c$ and conclude that $$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge\sqrt{36}=6$$

Last inequality was made so that $f(2,2,2)=g(2,2,2)$ and $f'_u(2,2,2)=g'_u(2,2,2)$, $f'_v(2,2,2)=g'_v(2,2,2)$, $f'_w(2,2,2)=g'_w(2,2,2)$ (and yes, we need to analyze $f''$ vs. $g''$ too). Last inequality DOES NOT hold when, say, $u=7,v=0,w=0$ (5.83 vs. 6.25) so one probably need another similar inequality to cover, say, cases like $u\approx3v\approx3w$

Note, that it may be possible that I shall not convert the sketch into real proof.

I do not want to say that this alternative approach (that closely mimics differentiation) is the only possible. There may be possible other approaches, for example, based on the following lemma:

Let $U\ge u \ge \delta\ge 0$, then $\sqrt{U}+\sqrt{u}\ge\sqrt{U+\delta}+\sqrt{u-\delta}$

Such approach may be much more elegant, but I still had no success with it.


Here is a sketch of another alternative approach.

Substitute variables $x=a+1,y=b+1,z=c+1$; then $$f(a,b,c)=\sqrt{4+a^2+bc+a}+\sqrt{4+b^2+ca+b}+\sqrt{4+c^2+ab+c}$$ — I exploited the fact that $a+b+c=x+y+z-3=0$; also note, that $-1\le a,b,c \le 2$

Use Taylor formula $\sqrt{4+t}=2\sqrt{1+\frac{t}{4}}=2+\frac{t}{4}-\frac{t^2}{16}+R(t)$ to get Taylor expansion for $f$ up to second degree; I got $$64f=64\cdot6+16(a+b+c)+15(a^2+b^2+c^2)+16(ab+bc+ca)+R(a,b,c)=64\cdot6+7(a^2+b^2+c^2)+R(a,b,c)$$ — I again exploit that $(a+b+c)^2=0$

So, it is local minimum; after getting some bound for $R(a,b,c)$ one may get a value for $\varepsilon$ to conclude that for any $a,b,c$ that $|a|<\varepsilon,|b|<\varepsilon,|c|<\varepsilon$ we have $f(a,b,c)\ge 6$

Next, we remember that $a^2+bc+a=x^2+yz\ge0,...$, so $$|f'_a(a,b,c)|\le |\frac{2a+1}{2\sqrt{2}}+\frac{c}{2\sqrt{2}}+\frac{b}{2\sqrt{2}}|\le\frac{9}{2\sqrt{2}}$$ and so $|f'_b(a,b,c)|\le \frac{9}{2\sqrt{2}}$, $|f'_c(a,b,c)|\le \frac{9}{2\sqrt{2}}$

Last step is just brute force calculation of values of $f$ in network of points $a=\frac{k}{N}$, $b=\frac{l}{N}$, $c=-\frac{k+l}{N}$ (except such values that lies in $|a|<\varepsilon,|b|<\varepsilon,|c|<\varepsilon$ where $f\approx 6$) where $N$ sufficiently large and $-N\le (k,l,-k-l)\le 2N$; after such calculation one may observe that $f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})>6+\frac{3\cdot 9}{2\sqrt{2}N}$ so there is no possibility that $f=6$, because $f(a,b,c)\ge f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})-\frac{1}{N}\frac{9}{2\sqrt{2}}(|a-\frac{k}{N}|+|b-\frac{l}{N}|+|c-\frac{-k-l}{N}|)>6$

If current $N$ is not enough to observe that $f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})>6+\frac{3\cdot 9}{2\sqrt{2}N}$, then one have to do brute force calculation with, say $N'=2N$; this will lead to success very soon, because $f''$ is small. I suspect that such calculation may be limited to <100 values, if one get good estimates for $R$.


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