<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] Just a remark :
On $x\in [1/2,2]$ we have the inequality :
$$f(x)=\left(1+x\right)^{\frac{1}{x}}\leq \frac{4}{x^{-1+\ln\left(4\right)}+1}$$
Adding the two quantities gives :
$$f(x)+f(1/x)\leq g(x)+g(1/x)=4$$
To show it we have the inequalities on $[1,2]$ the hardest part :
$$\left(1+x\right)^{\frac{1}{x}}\leq 2e^{\left(-\frac{5\left(-1+\ln\left(4\right)\right)}{2}\left(x^{\frac{1}{5}}-1\right)\right)}\leq 2e^{\left(1-x^{\frac{\left(-1+\ln\left(4\right)\right)}{2}}\right)}\leq \frac{4}{x^{\left(-1+\ln\left(4\right)\right)}+1}$$
For the fisrt we can substitute $y^5=x$
It's a bit long but fairly easy with logarithmic derivative .We cannot do simpler for the derivative of $f(x)$ .
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