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Prove that

$$ \exp\left(\frac{x+y}{2}\right)\le \exp(x)+\exp(y)+\exp(x+y), $$ for all $x,y\in\mathbb{R}$. There an algebraic approach to prove this inequality. Maybe some calculus manipulations will lead to this inequality. I wonder if it is possible to avoid derivative or another analysis stuffs. My attempt was to factor the expresiion obtained after putting all terms on the right-hand side.

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2 Answers 2

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Convexity: $$ \exp\left(\frac{x+y}{2}\right) \leq \frac{1}{2}\exp(x) + \frac{1}{2}\exp(y) \leq \exp(x) + \exp(y) \leq \exp(x) + \exp(y) + \exp(x+y) $$ So your estimation is actually way to brutal.

Or more elementary: Use Young's inequality, i.e. $uv \leq \frac{u^2+v^2}{2}$ for real numbers $u, v$ (comes directly from $0\leq (u-v)^2$, just expand). It follows: $$ \exp\left(\frac{x+y}{2}\right) = \exp\left(\frac{x}{2}\right) \exp\left(\frac{y}{2}\right)\leq \frac{\exp\left(\frac{x}{2}\right)^2+\exp\left(\frac{y}{2}\right)^2}{2} = \frac{\exp(x)+\exp(y)}{2} $$

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  • $\begingroup$ Interesting application of Young inequality. I was hoping that this particular case would lead me to the solution of a general inequality that I have to prove. It is about $ \exp\left( \left( 1-\lambda\right) x+\lambda y\right) \leq\left( 1+e^{x}\right) ^{1-\lambda}\left( 1+e^{y}\right) ^{\lambda}-1$, for $x,y\in\mathbb{R}$. It is possible to use Young inequality even in this general case? $\endgroup$
    – user217519
    Feb 3, 2022 at 22:45
  • $\begingroup$ For your follow-up question, use Jensen's inequality for the convex function $\log(1+e^x)$. $\endgroup$ Feb 3, 2022 at 23:14
  • $\begingroup$ @TannySieben is right. But first get $-1$ to the other side and apply $\log$ to both sides. I don't really think that Young will get you anywhere. $\endgroup$ Feb 3, 2022 at 23:17
  • $\begingroup$ @Tanny Sieben Yes, you are right. But there is a way that I can prove that $y=\ln(1+e^x)$ is a convex function without using derivatives? $\endgroup$
    – user217519
    Feb 4, 2022 at 7:58
  • $\begingroup$ Pretty much impossible... Is it your task to not use derivatives? $\endgroup$ Feb 5, 2022 at 13:31
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Let $a=e^\frac x2$ and $b=e^\frac y2$, so that your desired inequality is $ab \leq a^2+b^2+a^2b^2$ for $a, b>0$. This is easy to show, e.g. by $$ ab \leq \frac{a^2+b^2}{2} < a^2+b^2+a^2b^2 $$

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