Exponential of Average Less than Average of Exponential I am aiming to show the following inequality:
$$\exp\left[\frac{1}{b-a}\int_{a}^bf(x)dx\right] \leq \frac{1}{b-a}\int_a^b \exp[f(x)]dx$$
where $f(x) \in C([a,b])$.
Intuitively, this makes sense since for some real numbers $r$ and $s$
$$\exp\left[\frac{r}{2}+\frac{s}{2}\right] = \exp\left[\frac{r}{2}\right]\exp\left[\frac{s}{2}\right] = \sqrt{\exp[r]\exp[s]} \leq \frac{\exp[r]}{2} + \frac{\exp[s]}{2}$$
i.e. the arithmetic mean of two numbers is at least as much as the geometric mean (e.g. geometric mean of 1 and 9 is 3 while the arithmetic mean of 1 and 9 is 5). But how do I deal with a more generalized version of this statement dealing with integrals as expressed above?
 A: Arithmetic Geometric mean inequality states
$\dfrac 1 {b-a} \cdot \int_{a}^{b} g(x)dx\ge e^{\frac 1 {b-a} \cdot \int_{a}^b ln(g(x)) dx}$
for $g(x)$ being positive integrable function on $[a,b]$
To obtain your problem just replace $g(x)=e^{f(x)}$
A: I think you are on the right track. Consider 
$$ \sum_{i=1}^nf(x_i)\frac{b-a}{n}$$
where $x_i \in[a+(b-a)(i-1)/n,a+(b-a)i/n]$, is Riemann sum of $f$ continuous in $[a,b]$.  Then for positive $n$, and because of the geometric mean being less than the arithmetic mean,
\begin{align} 
\exp\left(\frac{1}{b-a}\sum_{i=1}^nf(x_i)\frac{b-a}{n}\right)&=\exp\left(\frac{1}{n}\sum_{i=1}^nf(x_i)\right)\\
&=\sqrt[n]{\prod_{i=1}^n\exp\left(f(x_i)\right)}\\
&\le \frac{1}{n}\sum_{i=1}^n\exp(f(x_i))\\
&=\frac{1}{b-a}\sum_{i=1}^n\exp(f(x_i))\frac{b-a}{n}
\end{align}
Take limits on both sides $n\rightarrow\infty$ and you get the result since the right side is Riemann sum of the function $\exp(f(\cdot))$. All continuous functions in a bounded closed interval are Riemann integrable.
