# Exponential inequality $\exp(\frac{x+y}{2})\le \exp(x)+\exp(y)+\exp(x+y)$

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.

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: $$\exp\left(\frac{x+y}{2}\right) = \exp\left(\frac{x}{2}\right) \exp\left(\frac{y}{2}\right)\leq \frac{\exp\left(\frac{x}{2}\right)^2+\exp\left(\frac{y}{2}\right)^2}{2} = \frac{\exp(x)+\exp(y)}{2}$$
• 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? Feb 3, 2022 at 22:45
• For your follow-up question, use Jensen's inequality for the convex function $\log(1+e^x)$. Feb 3, 2022 at 23:14
• @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. Feb 3, 2022 at 23:17
• @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? Feb 4, 2022 at 7:58
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 $$ab \leq \frac{a^2+b^2}{2} < a^2+b^2+a^2b^2$$