# An inequality with definite integrals

I need to prove that the following inequality holds:

$$\int_{0}^{e} \sqrt{e^x-1} + \int_{0}^{e} \log{(x^2+1)} \geq e^2$$

Any support is welcome. Thanks.

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Hint: Use Young's Inequality. –  David Mitra Jun 3 '12 at 19:36
@David Mitra: OK. Good idea! Thanks. –  Chris's sis Jun 3 '12 at 19:37
@DavidMitra Thats a really good hint hitting the nail on the head. Nice! I think you should add that as answer so that the question gets an answer. –  user17762 Jun 3 '12 at 19:40

Note that the inverse function of $f(x)=\sqrt{e^x-1}$, $0\le x\le e$, is $f^{-1}(x)=\log(x^2+1)$. So, you may apply the appropriate version of Young's Inequality.