TITLE: An application of Jensen's Inequality
QUESTION [10 upvotes]: Given that $\{\phi_n\}$ is a sequence of non-negative numbers whose sum is $1$ and $\{\psi_n\}$ is a sequence of positive numbers, how can I show that 
$$ \prod_{n=1}^{\infty}~\psi_n^{\phi_n}~\leq~\sum_{n=1}^{\infty}~\phi_n\psi_n~? $$  
Thanks.  
PS: I'm not too sure about the title. Perhaps, someone could give it a better title?

REPLY [3 votes]: We can apply Jensen inequality for a finite sums, i.e. $\sum_{k=1}^N\alpha_kf(x_k)\leq f\left(\sum_{k=1}^n\alpha_kx_k\right)$ if $\sum_{k=1}^n\alpha_k=1$. To see that, fix $n\in\mathbb N$ and put $c_k:=\frac{a_k}{\sum_{j=1}^na_j}$. Then by concavity of $x\mapsto \ln x$:
\begin{align}
\ln\prod_{k=1}^nb_k^{a_k}&=\sum_{k=1}^na_k\ln b_k\\
&=\left(\sum_{j=1}^na_j\right)\sum_{k=1}^nc_k\ln b_k\\
&\leq \left(\sum_{j=1}^na_j\right)\ln\left(\sum_{k=1}^nc_k b_k\right)
\\
&=\left(\sum_{j=1}^na_j\right)\left(\ln\left(\sum_{k=1}^na_k b_k\right)-\ln\left(\sum_{j=1}^na_j\right)\right)\\
&\leq\ln\left(\sum_{k=1}^{+\infty}a_k b_k\right)-\ln\left(\sum_{j=1}^na_j\right), 
\end{align}
and since $\lim_{n\to\infty}\sum_{j=1}^na_j=1$ we get the result, taking the $\limsup$.