<s>[INST] Prove that $(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ Prove that $f(x)=(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ for all $x>0.$
We have $f(x)=f(\frac{1}{x}), f'(x)=-\frac{1}{x^2}f'(\frac{1}{x}),$ so we only need to prove $f'(x)＞0$ for $0 < x < 1.$
 [/INST] From a friend who doesn't want to use math.se...
Write the inequality as
$$
\frac{1}{1+x} \left(1+x\right)^{1+\frac{1}{x}}  + \frac{x}{1+x} \left(1+\frac{1}{x}\right)^{1+x} \leq 4.$$
Let $f(x) =(1+x)^{1+\frac{1}{x}}$.  If $f$ is concave, then we are done since this means that 
$$4=f(1)=f((1- \alpha)x + \alpha y)) \geq (1- \alpha) f(x) + \alpha f(y) = \frac{1}{1+x} \left(1+x\right)^{1+\frac{1}{x}}  + \frac{x}{1+x} \left(1+\frac{1}{x}\right)^{1+x}$$ with $\alpha = \frac{x}{1+x}, 1-\alpha = \frac{1}{1+x},$ and $y = \frac{1}{x}$.  So, we will show that $f$ is concave.
Note that if $f(x) = e^{g(x)}$ and $g''(x) + g'(x)^2 \leq 0$ for all $x$, then $f$ is concave.
So, consider $g(x) = (1+\frac{1}{x}) \log (1+x).$  Then we have that
$$g''(x) + g'(x)^2 \leq 0 \leftrightarrow \log(1+x) \leq \frac{x}{\sqrt{1+x}}$$
Note that $\lim_{x \to 0^+} \log(1+x) = \lim_{x \to 0^+} \frac{x}{\sqrt{1+x}}=0.$
By AM-GM, 
$$\frac{1}{1+x} = \frac{d}{dx} \log(1+x) \leq \frac{x+2}{2(1+x)^{\frac{3}{2}}} = \frac{d}{dx} \frac{x}{\sqrt{1+x}} $$
So, $$\log(1+x) = \int_0^x \log (1+x) ' dx \leq \int_0^x \left(\frac{x}{\sqrt{1+x}}\right)' dx = \frac{x}{\sqrt{1+x}}.$$
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