Exponential of a sum comparing to a sum of exponentials (2 variables)

Question

I would like to prove for all $x, y \in \mathbb{R}$ that $\dfrac{e^{x}+e^{y}}{2} \geq e^{\frac{x+y}{2}}$. My idea, is to show that $f(x,y) \ge 0$, it means that $(0,0)$ is the minimum of $f(x,y)$. So, I compute the equation: $\nabla f(x,y)=\begin{pmatrix}0 \\0 \end{pmatrix}$. I find that the solutions are $x=y$. My first question is may I choose $\textbf{x=y=0}$? After I have made this assumption, I compute the eigenvalues of $\nabla^2f(x,y)_{(0,0)}$ and got $\lambda_1=0$ and $\lambda_2=\dfrac{1}{2}$. Thus, I can't conclude anything about $(0,0)$ from this point since one of the eigenvalues is zero. Do you have any idea about what can I do further using this method or a different path to prove it? Thank you.

Answer 1

$0 \le (e^{x/2}-e^{y/2})^2=e^x-2e^{\frac{x+y}{2}}+e^y$

Answer 2

Setting $e^x=u,e^y=v$, this is just

$$\frac{u+v}2\ge\sqrt{uv}.$$

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Dividing by $e^x$ and setting $t:=y-x$,

$$\frac{1+e^t}2\ge e^{t/2}.$$

$$1+\frac t2+\frac{t^2}{2\,2!}+\frac{t^3}{2\,3!}+\cdots\ge1+\frac t2+\frac{t^2}{2^22!}+\frac{t^3}{2^33!}+\cdots$$