Answer to Question #103131 in Algebra for ea

Question

Prove that
((x+y+z)/3)^x+y+z ≤ x^x + y^y + z^z ≤ ((x^2 + y^2 + z^2)/(x+y+z) ,
where x, y, z belongs to natural numbers.

Answer 1

I think there are some mistakes in the task.

For example, if $x=y=z=2$:

$((x+y+z)/3)^{x+y+z}=2^6\geq 3\times 2^2=x^x+y^y+z^z\geq(x^2+y^2+z^2)/(x+y+z)=2.$

But there should be another inequality signs.

To my mind, the correct task is as follows.

Prove for natural numbers x, y, z:

$\left( \frac{x+y+z}{3} \right)^{x+y+z}\leq x^x y^y z^z \leq \left( \frac{x^2+y^2+z^2}{x+y+z} \right)^{x+y+z}$

Proof:

1) Using inequality between arithmetic and geometric means for $a_1=x,...,a_ x=x, a_{x+1}=y,...,a_{x+y}=y,a_{x+y+1}=z,...,a_{x+y+z}=z,$

we have:

$x^x y^yz^z=a_1a_2...a_{x+y+z} \leq \left( \frac{a_1+a_2+...+a_{x+y+z}}{x+y+z} \right)^{x+y+z}= \left( \frac{x^2+y^2+z^2}{x+y+z} \right)^{x+y+z}$

2) Using inequality between arithmetic and geometric means for

$a_1=1/x,...,a_ x=1/x, a_{x+1}=1/y,...,a_{x+y}=1/y,a_{x+y+1}=1/z,...,a_{x+y+z}=1/z,$

we have:

$\left( \frac{3}{x+y+z} \right)^{x+y+z}= \left( \frac{\frac{1}{x}+...+\frac{1}{x}+\frac{1}{y}+...+\frac{1}{y}+\frac{1}{z}+...+\frac{1}{z}}{x+y+z} \right)^{x+y+z}\geq \left( \frac{1}{x} \right)^{x} \left( \frac{1}{y} \right)^{y} \left( \frac{1}{z} \right)^{z}$

$\left( \frac{3}{x+y+z} \right)^{x+y+z}\geq \left( \frac{1}{x} \right)^{x} \left( \frac{1}{y} \right)^{y} \left( \frac{1}{z} \right)^{z} \Rightarrow x^xy^yz^z\geq \left( \frac{x+y+z}{3} \right)^{x+y+z}$