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? enter image description here

up vote 6 down vote accepted

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 $$2^{e^x}-e^x\ln2-1=e^t-1-t>0.$$

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.