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November 3rd, 2017, 09:37 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27  Inequality proof
$\displaystyle x , y , z \in R^{+} \; $ are positive real numbers and $\displaystyle \; x+y+z=3$ prove $\displaystyle \; x^{y+z}y^{z+x}z^{x+y}\leq 1$ 
November 3rd, 2017, 12:40 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 415 Thanks: 228 Math Focus: Dynamical systems, analytic function theory, numerics 
Here is a rough sketch for you to fill in the details. 1. Prove that $x^{y+z}y^{x+z}z^{x+y} \leq (xyz)^3$. Hint: Use the weighted AM/GM inequality: \[ \left( \prod_{j=1}^n a_j^{w_j} \right)^{\frac{1}{W}} \leq \frac{1}{W} \sum_{j = 1}^{n} w_ja_j \] where $W = \sum_{j = 1}^{n} w_j$ where $a_j$ are each nonnegative. 2. Let $f(x,y,z) = (xyz)^3$ and let $D = \{(x,y,z) : x + y + z =1, x,y,z \geq 0 \}$. $f$ is continuous which means $D$ is closed so $f$ attains its maximum on $D$. Note that $f = 0$ on $\partial D$ so the inequality is satisfied there. On the interior one finds maxima of $f$ satisfying $x + y + z = 1$ only if $\nabla f$ is parallel to $(1,1,1)$. This is equivalent to $xy = yz = xz$ which is only satisfied when $x = y = z = 1$ and we see that in this case $x^{y+z}y^{x+z}z^{x+y} = 1$. Hence, the inequality holds for all $(x,y,z) \in D$ and in particular, on the interior where $x + y + z = 1$ and $x,y,z$ are positive. Alternatively, another application of AM/GM gives you immediately that \[(xyz)^3 \leq (\frac{x + y + z}{3})^3 = 1\] Last edited by SDK; November 3rd, 2017 at 12:55 PM. 
November 5th, 2017, 01:16 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27 
I post quick method , $\displaystyle y=(3x)lnx$ $\displaystyle y''<0 \Rightarrow y(a)+y(b)+y(c) \leq 3y(\frac{a+b+c}{3})$ $\displaystyle \ln a^{3a} + \ln b^{3b} +\ln c^{3c} \leq 3y(\frac{a+b+c}{3})=1$ $\displaystyle \ln a^{3a}b^{3b}c^{3c}\leq 1$ $\displaystyle a^{3a}b^{3b}c^{3c}=a^{b+c}b^{c+a}c^{a+b}\leq 1$ 

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