<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] Remarks: @Alvin Jin introduced a very nice proof. Here I give an alternative proof.
The idea is to find a function $f(x)$ such that $f(x) + f(1/x) \equiv 4$ and
$\left(1 + \frac{1}{x}\right)^x \le f(x)$ for all $x > 0$.
Then we have $\left(1 + x\right)^{1/x} \le f(1/x)$ for all $x > 0$.
Adding them up, we have $\left(1 + \frac{1}{x}\right)^x + \left(1 + x\right)^{1/x} \le f(x) + f(1/x) = 4$.

Fact 1: It holds that, for all $x > 0$,
$$\left(1 + \frac{1}{x}\right)^x
\le \frac{4\ln 2 \cdot x + 4 - 4\ln 2}{x + 1}. \tag{1}$$
(The proof is given at the end.)
By Fact 1 (1), letting $x \to \frac{1}{x}$, we have, for all $x > 0$,
$$\left(1 + x\right)^{1/x}
\le \frac{4\ln 2 \cdot \frac{1}{x} + 4 - 4\ln 2}{\frac{1}{x} + 1} = \frac{4\ln 2  + (4 - 4\ln 2)x}{1 + x}. \tag{2}$$
$(1) + (2)$ gives
$$\left(1 + \frac{1}{x}\right)^x + \left(1 + x\right)^{1/x}
\le 4.$$
We are done.

Proof of Fact 1:
It suffice to prove that
$$\frac{(1 + x)^{1 + x}}{x^x} \le 4\ln 2 \cdot x + 4 - 4\ln 2.$$
Taking logarithm, it suffices to prove that
$$(1 + x)\ln (1 + x) - x\ln x \le \ln (4\ln 2 \cdot x + 4 - 4\ln 2).$$
Let $g(x) := \mathrm{LHS} - \mathrm{RHS}$. We have
$$g'(x) = \ln(1 + x) - \ln x - \frac{1}{x + \frac{1}{\ln 2} - 1}$$
and
$$g''(x) = \frac{(1-2a)x - a^2}{x(1+x)(x+a)^2}$$
where $a = \frac{1}{\ln 2} - 1 \approx 0.4427$.
Let $x_0 = \frac{a^2}{1-2a} \approx 1.7$. We have
$g''(x) < 0$ on $(0, x_0)$,
and $g''(x) > 0$ on $(x_0, \infty)$,
and $g''(x_0) = 0$.
Note that $g'(1) = 0$ and $\lim_{x\to \infty} g'(x) = 0$.
Thus, we have $g'(x) > 0$ on $(0, 1)$,
and $g'(x) < 0$ on $(1, \infty)$.
Thus, we have $g(x) \le g(1) = 0$ on $(0, \infty)$.
We are done.
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