Partial solution: We must prove that the function $f(x)={{2}^{-x}}+{{2}^{-\frac{1}{x}}}-1$ defined for $x>0$, have a maximum in $x=1$. We calculate the derivative, and we get: $f'(x)=\ln 2\left( \frac{{{2}^{-\frac{1}{x}}}}{{{x}^{2}}}-{{2}^{-x}} \right)$. This function have a root $x=1$. We can prove that $f'(x)>0$, for $x>1$. Indeed, we must prove that ${{2}^{x-\frac{1}{x}}}<{{x}^{2}}$, for $x>1$. We consider the function $g(x)=\ln x-a\left( x-\frac{1}{x} \right),\text{ }a=\ln \sqrt{2}$. Using the derivative we can see that this function it is monotonous increasing, so we have $g(x)>g(1)=0$, so in $x=1$ it is a maximum point fot the function $f$, so we have $f(x)<=f(1)=0$. But we must analyze what's happen which the sign of the function $f'(x), i.e. g(x)$ in the interval $(0,1)$, but seems that this is a little bit difficult, see tha picture.

I can't believe that (...) https://artofproblemsolving.com/community/c6h1598849p9938551.

Filipjack wrote: I can't believe that (...) https://artofproblemsolving.com/community/c6h1598849p9938551. Yes, and I heard that it is even older than that, from a test in 2012.

So the problem it is not easy at all..

The proofs at that link all appear to be overcomplicating the problem.

An alternative solution (in Chinese) can be seen here.

Here's a solution that doesn't involve computing a nasty derivative in $x$. (At least, if this works, which I think it does? It's a bit jank, though.) Let $f(x) = 2^{-x} + 2^{-1/x}$; remark that $f(x) = f(\tfrac 1x)$, so it suffices to examine $x$ in the range $(0,1]$. Define $M := \sup_{x\in\mathbb (0,1]}f(x)$. At least one of the following two cases must occur. Case 1: The supremum is achieved as $\color{blue}x\to 0$. Compute $\lim_{x\to 0^+}f(x) = 1 + 0 = 1$, so $M = 1$ in this case. Case 2: the supremum is achieved for some $\color{blue} x_0\in (0,1]$. Choose some interval $I = [\tfrac{x_0}2, 1+\varepsilon]$ containing $x_0$, so the supremum is achieved strictly in the interior of $I$. Now use Lagrange Multipliers. Let $u:= 2^{-x}$ and $v := 2^{-1/x}$. The quantities $u$ and $v$ are related by the equation $(\log u)(\log v) = (\log 2)^2$, where all logarithms are taken in base $e$. We thus seek to maximize $g(u,v) := u + v$ subject to $0 = h(u,v) := (\log u)(\log v) - (\log 2)^2$. Compute \[ \nabla g(u,v) = \langle v,u\rangle \quad\text{and}\quad \nabla h(u,v) = \left\langle \frac{\log v}{u}, \frac{\log u}{v}\right\rangle. \]Then the condition of Lagrange Multipliers states there exists $\lambda \in \mathbb R$ such that \[ \frac{\log v}{u} = \lambda v\quad\text{and}\quad \frac{\log u}{v} = \lambda u. \]This implies $\log v = \lambda uv = \log u$, so $v = u$. This occurs when $x_0 = 1$, which gives $f(1) = 2^{-1} + 2^{-1} = 1$. Thus $M = 1$ in both cases, which concludes the proof. $\blacksquare$