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

Find ln through the formula:

(x+y+z)ln((x+y+z)/3)$\leq$ xlnx+ylny+zlnz $\leq$ (x+y+z)ln((x^2+y^2+z^2)/(x+y+z))

Let's prove the left-hand side:

Let f(x)=xlnx

f'(x)=lnx+1.

Hence, f′′(x)=1/x>0, which says that f is a convex function.

Thus, by Jensen inequality,

(xlnx+ylny+zlnz)/3 $\geq$ (x+y+z)/3 * ln((x+y+z)/3),

so, xlnx+ylny+zlnz $\geq$ (x+y+z) * ln((x+y+z)/3).

Let's prove the right-hand side:

xlnx+ylny+zlnz $\leq$ (x+y+z)ln((x²+y²+z²)/(x+y+z));

(x+y+z)ln((x²+y²+z²)/(x+y+z))- (xlnx+ylny+zlnz) $\geq$ 0;

xln((x²+y²+z²)/(x+y+z))+yln((x²+y²+z²)/(x+y+z))+zln((x²+y²+z²)/(x+y+z))- (xlnx+ylny+zlnz) $\geq$ 0,

xln((x²+y²+z²)/(x*(x+y+z)))+yln((x²+y²+z²)/(y*(x+y+z)))+zln((x²+y²+z²)/(z*(x+y+z))) $\geq$ 0

because it can be proved that (x²+y²+z²)/(x*(x+y+z))$\geq 1$ , (x²+y²+z²)/(y*(x+y+z)) $\geq$ 1, (x²+y²+z²)/(z*(x+y+z)) $\geq$ 1 and ln((x²+y²+z²)/(x*(x+y+z))) $\geq$ 0,

ln((x²+y²+z²)/(y*(x+y+z))) $\geq$ 0, ln((x²+y²+z²)/(z*(x+y+z))) $\geq$ 0.

So we have the sum of three non-negative terms, it also will be non-negative.

Thus, the inequalities were proved.