Your inequality $\prod_{n=1}^{\infty}~\psi_n^{\phi_n}~\leq~\sum_{n=1}^{\infty}~\phi_n\psi_n$ is equivalent to $\sum_{n=1}^{\infty} {\phi_n} \ln \psi_n \leq \ln \left(\sum_{n=1}^{\infty} \phi_n\psi_n\right).$
This can be obtained from the integral form of Jensen's inequality: The following is true for any concave function $\varphi\left(\int_\Omega g\, d\mu\right) \ge \int_\Omega \varphi \circ g\, d\mu. $ (I have copied it from wikipedia, just changed convex to concave and change the inequality sign.)
You obtain the above inequality for:
- concave function $\varphi(x)=\ln(x)$
- measure space $\Omega=\mathbb N$ with the measure determined by $\mu(\{n\})=\phi_n$
- function $g:\Omega\to\mathbb R$ given by $g(n)=\psi_n$.
The measure $\mu$ is sometimes called discrete measure.
Or, equivalently, you can look at $\psi_n$'s as probabilities and then use the probabilistic form of Jensen's inequality. (Comparing the mean value of logarithm and logarithm of the mean value.)
If you prefer, you might also use the finite version of Jensen's inequality: $\varphi\left(\frac{\sum_{i=1}^k a_i x_i}{\sum_{j=1}^k a_j}\right) \ge \frac{\sum_{i=1}^k a_i \varphi (x_i)}{\sum_{j=1}^k a_j}$ (again - the above formula is copied from wikipedia and I changed the sign) and then take the limit.
Alternatively, you can use the finite version to prove $\prod_{n=1}^{N} \psi_n^{\phi_n} \leq \sum_{n=1}^{N} \phi_n\psi_n \qquad (*)$ for arbitrary finite sequence $(\phi_n)$ fulfilling $\sum_{n=1}^{N} \phi_n=1$, use this for sequence and then use this to derive the same thing for your original -- infinite -- sequence, by taking the inequality (*) for $\phi_1,\dots,\phi_N,1-\sum_{i=1}^N \phi_i$ and $\psi_1,\dots,\psi_n,0$. (In this way the limits are slightly less complicated and you do not have the sum $\sum_{j=1}^k \phi_j$ in denominator.)
BTW It's worth mentioning that this is a generalization of AM-GM inequality.