Prove that $e^{\frac{x+y}{2}} \le \frac{e^x + e^y}{2}$

Question

To prove: $e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2}$

We observe:

$e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2} $

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le 1/2\sum_{n=0}^\infty \dfrac{x^n}{n!} + 1/2\sum_{n=0}^\infty \dfrac{y^n}{n!}$

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty 1/2\dfrac{x^n}{n!} + \sum_{n=0}^\infty 1/2 \dfrac{y^n}{n!} $

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty \dfrac{x^n + y^n}{2\cdot n!}$

In essence, now we need to show that $ \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \dfrac{x^n + y^n}{2\cdot n!}$

I'm looking for an idea or a hint to help me solve this. I'm thinking about using something from the binomial theorem. Maybe I made a mistake as well.

Answer 1

Another solution :

$e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2}$ comes directly from the convexity of $x\mapsto e^x$.

A convex function is a function such that for all $x,y$, and $\lambda \in [0,1]$, $f(\lambda x + (1-\lambda)y)\leqslant \lambda f(x) + (1-\lambda) f(y)$.

With $\lambda = 1/2$ you get your answer.

(and you can prove exp 's convexity, by deriving it two times)

Answer 2

Let $y=x+t$ then we have to prove $$e^{x+t/2}\le \frac{e^x+e^{x+t}}{2}$$ or $$\frac{e^t+1}{2}\ge e^{t/2}$$ $$\iff {(e^{t/2}-1)}^2\ge 0$$ which is true

Answer 3

$$\frac{e^x+e^y}{2}=e^{\frac{x+y}{2}} \cdot \frac{e^{\frac{x-y}{2}}+e^{\frac{y-x}{2}}}{2}$$ then, $$\frac{e^x+e^y}{2}=e^{\frac{x+y}{2}} \cdot \cosh\left ( e^{\frac{x-y}{2}} \right )$$

as $\forall z, \cosh(z) \geq 1$, then

$$\frac{e^x+e^y}{2} \geq e^{\frac{x+y}{2}}$$

Answer 4

It is sufficient to notice that :

$e^{\frac{x+y}{2}}= \sqrt{e^xe^y}$

The Inequality of arithmetic and geometric(AM-GM) means is:

$ \sqrt{ab}\le \frac{a+b}{2}$

Let $a=e^x, b=e^y$. Then we are done!

Answer 5

One more proof: define $f_y(x):=e^x+e^y-2e^{(x+y)/2}$ at fixed $y$ so $f_y(y)=0$ while $f_y^\prime(x)=e^x-e^{(x+y)/2}$. Since $f_y^\prime(x)$ has the same sign as $x-y$, $x=y$ is the global minimum.