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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.
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: $$ \exp\left(\frac{x+y}{2}\right) = \exp\left(\frac{x}{2}\right) \exp\left(\frac{y}{2}\right)\leq \frac{\exp\left(\frac{x}{2}\right)^2+\exp\left(\frac{y}{2}\right)^2}{2} = \frac{\exp(x)+\exp(y)}{2} $$
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 $$ ab \leq \frac{a^2+b^2}{2} < a^2+b^2+a^2b^2 $$
