Proof; Log Convex implies Convexity I'm wondering if someone can check my work for me and provide any feedback. Cheers.

Problem: Suppose that $f:(a,b) \to (0,\infty)$ us such that $\log f$ is convex. Prove that
  $f$ is convex.

 A: Proof: We remind ourselves that a function $h(x)$ is convex if
\begin{align*} h( \lambda x + (1- \lambda)y) \leq \lambda h(x) + (1-\lambda) h(y)\end{align*}
for some $\lambda \in (0,1)$. Now, we know that $g(x) = \log f(x)$ is a convex function. Therefore 
\begin{align*} g( \lambda x + (1- \lambda)y) \leq \lambda g(x) + (1-\lambda) g(y),\end{align*}
which by definition gives
\begin{align*} \log f( \lambda x + (1- \lambda)y) \leq \lambda \log f(x) + (1-\lambda) \log f(y).\end{align*}
Since we know that the exponential is monotonically increasing we thus have
\begin{align*} f( \lambda x + (1- \lambda)y) \leq f(x)^\lambda f(y)^{1-\lambda}.\end{align*}
To prove the statement we utilize the following theorem

Theorem Let $a_1,...,a_n \geq 0$ and $p_1,...,p_n >0$ with $\sum p_i = 1$. Then
  \begin{align*} \prod_{i=1}^n a_i^{p_i} \leq \sum_{i=1}^n p_i a_i,
\end{align*}
  with equality if and only if $a_1=a_2=...=a_n$.

Thus, by the above theorem we have
\begin{align*} f( \lambda x + (1- \lambda)y) \leq \lambda f(x)+ (1-\lambda) f(y)\end{align*}
and we are therefore able to conclude that $f(x)$ is a convex function.
