Answer to Question #118022 in Algebra for benard

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))^(x+y+z) Where x, y, z element of N

Answer 1

To solve this problem, we will use

AM ≥GM twice.

Firstly, let us consider

x,x,. ... x-times, y,y,....... y-times,z,z.... z-times.

AM of above numbers is $\frac {x.x + y.y + z.z} {x+y+z}$ = $\frac {x² + y² + z²} {x+y+z}$

GM of above numbers = ($\ x^x . y^y . z^z$ )$\ ^ \frac {1} {(x+y+z)}$

Since AM ≥ GM,

$\frac {x²+y²+z²} {x+y+z}$ ≥ ($\ x^x . y^y . z^z$ )$\ ^ \frac {1} {(x+y+z)}$

=> ( $\frac {x²+y²+z²} {x+y+z}$ )$\ ^ {x+y+z}$ ≥ $\ x^x . y^y . z^z$ -----------> (1)

To prove  $(\frac{x+y+z}{3})^{x+y+z} \leq x^x y^y z^z$, secondly, let us consider

1/x, 1/x,. ... x-times, 1/y, 1/y,....... y-times,1/z, 1/z.... z-times.

AM of above numbers is $\frac {x.(1/x) + y.(1/y) + z.(1/z)} {x+y+z}$ = $\frac {3} {x+y+z}$

GM of above numbers =

$\ [ (1/x)^x . (1/y)^y . (1/z)^z ]$ $\ ^ \frac {1} {(x+y+z)}$

= [ $\frac {1} {x^xy^yz^z}$ ]$\ ^ \frac {1} {(x+y+z)}$

Since AM ≥ GM

$\frac {3} {x+y+z}$ ≥ [$\frac {1} {x^xy^yz^z}$ ]$\ ^ \frac {1} {(x+y+z)}$

=> ($\frac {3} {x+y+z}$ )^(x+y+z) ≥ $\frac {1} {x^xy^yz^z}$

Reciprocating,

$\ x^x . y^y . z^z$ ≥ ($\frac {x+y+z} {3}$ )^(x+y+z) ---------> (2)

By inequality (1) and (2)

($\frac {x²+y²+z²} {x+y+z}$ )^(x+y+z) ≥ $\ x^x . y^y . z^z$ ≥

( $\frac {x+y+z} {3}$ )^(x+y+z)

=> ($\frac {x+y+z} {3}$ )^(x+y+z)≤ $\ x^x . y^y . z^z$ ≤

($\frac {x²+y²+z²} {x+y+z}$ )^(x+y+z)