Let $x>0.$ Prove that $$2^{-x}+2^{-1/x} \leq 1.$$
Problem
Source: Romania NMO - 2018
Tags: calculus, inequalities
07.04.2018 11:30
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.
Attachments:

07.04.2018 12:03
I can't believe that (...) https://artofproblemsolving.com/community/c6h1598849p9938551.
07.04.2018 12:10
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.
07.04.2018 12:18
So the problem it is not easy at all..
07.04.2018 20:02
The proofs at that link all appear to be overcomplicating the problem.
06.05.2021 01:44
An alternative solution (in Chinese) can be seen here.
24.07.2022 05:48
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. $\blacksquare$