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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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    @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.2012-06-03

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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.