Prove: for all $x \in (0, 1], 2^x+2^{\frac{1}{x}} \leqslant 2^{x+\frac{1}{x}}$

Question

Prove: for all $x \in (0, 1], 2^x+2^{\frac{1}{x}} \leqslant 2^{x+\frac{1}{x}}$

This problem is from my math teacher. I tried using Calculus, the derivative function is like a black hole. Then I graphed it by Mathematica. As the following picture shows, I was strongly astonished. How can this inequality be proved?

Answer 1

Let $x=e^{a}$ and $\frac{1}{x}=e^{b}$.

Thus, $a+b=0$ and we need to prove that $$(2^x-1)\left(2^{\frac{1}{x}}-1\right)\geq1$$ or $$\ln\left(2^{e^a}-1\right)+\ln\left(2^{e^b}-1\right)\geq0,$$ which is just Jensen for $f(x)=\ln\left(2^{e^x}-1\right)$.

Indeed, $$f''(x)=\frac{2^{e^x}e^x\ln2\left(2^{e^x}-e^x\ln2-1\right)}{\left(2^{e^x}-1\right)^2}>0$$ because if $e^x\ln2=t$ then $t>0$ and $$2^{e^x}-e^x\ln2-1=e^t-1-t>0.$$