Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Prove:

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

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

share|improve this question
4  
Hint: How much is $(x+y+z)^2$ ? –  Lucian May 1 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 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 at 12:18
    
Calvin Lin use am gm inequalyty –  Math can be Fun May 1 at 12:24
2  
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 at 12:26
show 16 more comments

3 Answers 3

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.

share|improve this answer
    
Thanks a lot Ewan! –  FuriousMathematician May 1 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 at 12:55
add comment

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.

share|improve this answer
1  
Calvin is a genius –  Math can be Fun May 1 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 at 13:11
    
Ah, that's an approach for the toolbox then. Thanks. –  Macavity May 1 at 13:17
2  
@CalvinLin Adding the condition $x+y+z=1$, does it not change the given question? –  Mridul Sachdeva May 1 at 14:09
1  
@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 at 15:58
show 7 more comments

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}

share|improve this answer
1  
But $$\sqrt[3]{3}/3=\sqrt[3]{1/9}$$... –  Leverkuehn May 1 at 15:59
1  
-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 at 16:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.