$x,y \in R$ and $x \neq y$ Show that $e^\frac{x+y}2 \lt \frac12 (e^x + e^y)$

Question

Let $x,y \in R$ and $x \neq y$. Show that $e^\frac{x+y}2 \lt \frac12 (e^x + e^y)$

Struggling with this proof atm.

Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.

Answer 1

if $a\ne b$

$(a-b)^2 > 0\\ a^2 + b^2 > 2ab\\ ab < \frac 12 (a^2 + b^2)$

This is the AM-GM inequality.

$a = e^\frac x2, b= e^\frac y2$

Answer 2

By convexity for $f(x)=e^x$, that is by Jensen's inequality for $x\neq y$

$$f(\lambda x+(1-\lambda)y) < \lambda f(x)+(1-\lambda)f(y)$$

with $\lambda=\frac12$ we have

$$e^{\frac{x+y}2} < \frac12 (e^x + e^y)$$

Answer 3

We have $${(e^{x\over 2}-e^{y\over 2})^2>0\\e^x+e^y>2e^{x\over 2}e^{y\over 2}\\{1\over 2}(e^x+e^y)>e^{x+y\over 2}}$$