How can I prove the following exponential inequality?

Question

I'm trying to prove the following inequality with no success: $$ e^{-\frac{1}{x}}e^{-\frac{1}{y}}e^{-\frac{1}{x+y}} \leq e^{-\frac{1}{x}}+e^{-\frac{1}{y}}-e^{-\frac{1}{x+y}} $$ for $x>1$, and $y>1$. Does anybody have an idea on how to do this proof?

Answer 1

The inequality is equal to : $(e^x+e^{y})e^{\frac{xy}{x+y}}-e^{x+y} \ge 1 $ and $ 0\le x \le 1, 0\le y \le 1$

WLOG, let $x \ge y, y=tx, \implies 0 \le t \le 1 $,

let $u=e^x \implies 1 \le u \le e, $

$ \iff (u+u^t)u^{\frac{t}{t+1}} \ge u^{1+t}+1 \iff u^{1+\frac{t}{t+1}} +u^{t+\frac{t}{t+1}} \ge u^{1+t}+1$

since : $\dfrac{t}{t+1} \ge \dfrac{t}{2},t+\dfrac{t}{t+1} \ge \dfrac{3t}{2}$

LHS $\ge u^{1+\frac{t}{2}}+u^{\frac{3t}{2}} \ge (u^{1+t}+1) $

let $f(t)=u^{1+\frac{t}{2}}+u^{\frac{3t}{2}}-(u^{1+t}+1)$, we will prove $f(t)$ is mono increasing function:

$f'(t)=ln(u)(\dfrac{1}{2}u^{1+\frac{t}{2}}+\dfrac{3}{2}u^{\frac{3t}{2}}-u^{1+t})$, since $ln(u) \ge 0$ .it remains:

$u^{1+\frac{t}{2}}+3u^{\frac{3t}{2}}-2u^{1+t}>0 \iff 1+3u^{t-1} -2u^{\frac{t}{2}}>0$, let $g(t)=1+3u^{t-1} -2u^{\frac{t}{2}}$, we will prove $g(t)$ is mono increasing function also:

$g'(t)=ln(u)(3u^{t-1}-u^{\frac{t}{2}})>0 \iff 3u^{t-1}>u^{\frac{t}{2}} \iff 3 > u^{1-\frac{t}{2}}$

but $u^{1-\frac{t}{2}}$ is mono decreasing function, so max of $u^{1-\frac{t}{2}}$ is $u \le e <3 $

$g(0)=1+\dfrac{3}{e}-2>0 \implies f'(t)>0, f_{min}=f(0)=0 \implies f(t) \ge 0 $

QED.