$\dfrac{e^t + e^{-t}}{2} \leq e^{ \frac{t^2}{2} } $

Asked Apr 17 '21 at 16:36

Active Apr 17 '21 at 16:46

Viewed 114 times

We want to prove that $ \forall t, \dfrac{e^t + e^{-t}}{2} \leq e^{ \frac{t^2}{2} } $

I thought to a convexity inequality. If $\psi$ is convex, $\dfrac{ \psi(a)+\psi(b) }{2} \leq \psi( \frac{ a+b }{2} )$ or it may be the arithmetic-geometric inequality.

edited Apr 17 '21 at 16:41

asked Apr 17 '21 at 16:36

zestiria

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Duplicates found with Approach0 – Martin R Apr 17 '21 at 16:45
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The Maclaurin expansion might be of help – shsh23 Apr 17 '21 at 16:46
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Look here also https://math.stackexchange.com/questions/331367/cosh-x-inequality – framago Apr 17 '21 at 16:46
ok, thanks so much. – zestiria Apr 17 '21 at 16:46

$\dfrac{e^t + e^{-t}}{2} \leq e^{ \frac{t^2}{2} } $

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