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If: $x^2+y^2+z^2=2(xy+xz+zy)$ and $x,y,z \in R^+$


$\frac{x+y+z}{3} \ge \sqrt[3]{2xyz}$

I tried my best to solve this thing but no use. Hope you guys can help me.Thanks in advance.

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Hint: How much is $(x+y+z)^2$ ? – Lucian May 1 '14 at 11:59
@lucian can you elaborate on how to use that hint? Standard inequality techniques will not work since the equality case is not when all variables are equal. – Calvin Lin May 1 '14 at 12:13
So the hint is not an actual hint in the sense of leading towards a solution, but is just something nice that you noticed? – Calvin Lin May 1 '14 at 12:18
Calvin Lin use am gm inequalyty – Math can be Fun May 1 '14 at 12:24
So far I obtained $$\left( \frac{x+y+z}{3} \right )^2 = \frac 4 9 \left( xy + yz + zx \right ) \ge \frac 4 3 \left( xyz \right )^{\frac 2 3 }$$ still not good enough – Santosh Linkha May 1 '14 at 12:26
up vote 11 down vote accepted

If we put $a=\sqrt{x}$ and $b=\sqrt{y}$, the degree two equation (in $z$) $x^2+y^2+z^2-2(xy+xz+yz)=0$ has two solutions, $(a-b)^2$ and $(a+b)^2$. By cyclically permuting $x,y,z$, we may assume $z=(a+b)^2$. The inequality to be shown is then equivalent to $(x+y+z)^3 \geq 54xyz$, or $(a^2+b^2+(a+b)^2)^3 \geq 54(a^2b^2(a+b)^2)$. We are then done because

$$ (a^2+b^2+(a+b)^2)^3 -54(a^2b^2(a+b)^2)=2\Bigg((b-a)(2a+b)(a+2b)\Bigg)^2 $$

As guessed by CalvinLin, equality is reached exactly when $(x,y,z)=(1,1,4)$ up to permutation.

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Thanks a lot Ewan! – FuriousMathematician May 1 '14 at 12:52
By the way, the main identitiy is homogeneous so it is not too tedious to check it by hand. – Ewan Delanoy May 1 '14 at 12:55

As the equations are all homogenous, we'd add the condition that $x+y+z=1$. This gives us $ 1 = (x+y+z)^2 = 4 (xy + yz + zx) $. Let $C = xyz$, which is a positive number. We want to show that $ 0\leq C \leq \frac{1}{54}$.

Consider the cubic equation with roots $x, y, z$. It has the form $ X^3 - X^2 + \frac{1}{4} X - C$. For a cubic equation to have 3 real roots, it must have a non-negative discriminant, which gives us $ C-54C^2 \geq 0$ (courtesy of Wolfram, sorry I screwed this up earlier), or hence that $ 0\leq C \leq \frac{1}{54}$. Hence we are done.

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Calvin is a genius – Math can be Fun May 1 '14 at 13:07
@Macavity I've been using this technique to set problems, which is why it came to me. It seems to work well with cases where $a = b \neq c $, and the newton sums are easily determined. – Calvin Lin May 1 '14 at 13:11
Ah, that's an approach for the toolbox then. Thanks. – Macavity May 1 '14 at 13:17
@CalvinLin Adding the condition $x+y+z=1$, does it not change the given question? – idpd15 May 1 '14 at 14:09
@mathh apply vieta formula. We have x+y+z=1 and xy+Yz+zx=1/4 and xyz=1. Hence the cubic polynomial follows. – Calvin Lin May 1 '14 at 15:58

You have: \begin{align} (x + y + z)^2 &= x^2 + y^2 + z^2 + 2 (x y + x z + y z) \\ &= 4 (x y + x z + y z) \\ &= 4 \cdot 3 \cdot \frac{x y + x z + y z}{3} \\ &\ge 12 \sqrt[3]{x^2 y^2 z^2} \\ x + y + z &\ge 2 \sqrt[3]{3 x y z} \\ \frac{x + y + z}{3} &\ge \sqrt[3]{\frac{8}{9} x y z} \end{align}

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But $$\sqrt[3]{3}/3=\sqrt[3]{1/9}$$... – Leverkuehn May 1 '14 at 15:59
-1 please check your constants and cube / square roots. You made 2 errors. See Santosh comment up above for the correct value – Calvin Lin May 1 '14 at 16:12

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