What $\alpha$ such that if $xy=\alpha$, then $e^{-x}+e^{-y}\geq 2e^{-\sqrt \alpha} $?

Question

For every $ x,y \gt 0$, if $ xy=\alpha$, then we have

$$e^{-x}+e^{-y}\geq 2e^{-\sqrt \alpha} $$

What are the possible values of $\alpha$?

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$2 < e^{1/(n+1)} + e^{-1/n}$ led to this problem. so, $\alpha=1$ is allowed.

The problem is difficult, if not very: $e^{-x}+e^{-\frac1x}\geq 2e^{-1}$ is not easy to prove.

P.S.:
1) $\quad e^{-\sqrt \alpha}\geq e^{\dfrac{-x-y}2}$

2) $\quad e^{-x}+e^{-y}\geq 2e^{-\sqrt {xy}} $ does not hold for all $x,y\gt0$! Just think $x=1,y\to0$

Any help will be appreciated!

Answer 1

The argument in my answer to the linked question extends with minor modifications to the case $\alpha\ge 1$; thus if $\alpha\ge 1$ then the global minimum of $e^{-x}+e^{-y}$ on $xy=\alpha$ is also the only critical point, and occurs at $x=y=\sqrt\alpha$, so the minimum is indeed $2e^{-\sqrt\alpha}$.

On the other hand, if $0 < \alpha < 1$, then for $f(x)=e^{-x}+e^{-\alpha/x}$ we have, by direct computation, $$ f'(\sqrt\alpha) = 0 \qquad\text{and}\qquad f''(\sqrt\alpha) = 2\Big(1-\frac1{\sqrt\alpha}\Big) e^{-\sqrt\alpha} < 0 $$ Since $f''$ is continuous, this shows that the value of $f$ at $\sqrt\alpha$ is a strict local maximum. In particular, there are $x$ near $\sqrt\alpha$ such that $f(x)<f(\sqrt\alpha)=2e^{-\sqrt\alpha}$, so the inequality does not hold when $0<\alpha<1$.