If $x,y,z\in\mathbb R^+$ are in Harmonic Progression, prove that $z\cdot e^{x-y}+x\cdot e^{z-y}\ge\frac{2xz}y$

Question

If $x,y,z\in\mathbb R^+$ are in Harmonic Progression, prove that $z\cdot e^{x-y}+x\cdot e^{z-y}\ge\frac{2xz}y$

Given, $\frac1y-\frac1x=\frac1z-\frac1y\implies z(x-y)=x(y-z)$

I tried AM-GM inequality on $ze^{x-y}, xe^{z-y}$ but couldn't conclude.

I tried weighted AM-GM inequality on $e^{x-y}$ with weight $z$ and $e^{z-y}$ with weight $x$ but couldn't conclude.

I tried AM-HM inequality but couldn't conclude.

How to solve this problem?

Answer 1

Your inequality is equivalent to: $$\dfrac{e^x}{x} + \dfrac{e^z}{z}\geq2\dfrac{e^y}{y}\iff \dfrac{e^x}{x} + \dfrac{e^z}{z}\geq\left(\frac 1x + \frac 1z\right)e^{\tfrac{2xz}{x+z}}.$$ After some more rearrangement, this is equivalent to: $$\frac{z}{x+z}e^x + \frac{x}{x+z}e^z\geq e^{\tfrac{x\cdot z}{x+z} + \tfrac{z\cdot x}{x+z}},$$ which is just weighted AM-GM.

Interestingly, the Jensen's inequality is not strong enough in the first form as $\dfrac{e^x}{x}$ is convex but not increasing.

Answer 2

As @dezdichado said, the inequality can be rewritten as $$ \frac 1y e^y \le \frac 12 \left( \frac 1x e^x + \frac 1z e^z\right) \, . $$ With the substitution $x=1/u$, $y=1/v$, $z=1/w$ this becomes $$ v e^{1/v} \le \frac 12 \left( u e^{1/u} + w e^{1/w}\right) $$ with $v = (u+w)/2$, this is Jensen's inequality for the convex function $f(x) = x e^{1/x}$.