TITLE: Show that $e^{f(x)}$ is convex.
QUESTION [13 upvotes]: Let $f : (0,\infty) \to \mathbb{R}$ be a convex function. Prove that $e^{f(x)}$ is a convex function on $(0,\infty)$. 

My original idea was to try and show that the second derivative is positive, but this will not work since $f(x)$ need not be differentiable. Here's my second attempt:
By definition, since $f$ is convex, we have that $f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$ for any $\lambda \in (0,1)$ and $x,y \in (0,\infty)$. Then, by applying the exponential to both sides we have that $e^{f(\lambda x+(1-\lambda)y))}\leq e^{\lambda f(x)+(1-\lambda)f(y)}$ . Applying rules of exponents, $e^{f(\lambda x+(1-\lambda)y))} \leq e^{\lambda f(x)}e^{(1-\lambda)f(y)}$. From here, I want to bring the $\lambda$ and $1-\lambda$ terms down in front of the exponential, but I am stuck as to how to do this.

REPLY [5 votes]: "From here, I want to bring the $\lambda$ and $1−λ$ terms down in front of the exponential, but I am stuck as to how to do this."
If you wish to continue from where you stuck: denoting $e^{f(x)}=A$, $e^{f(y)}=B$, you have to prove that 
$$
A^\lambda B^{1-\lambda}\le\lambda A+(1-\lambda)B.
$$
Taking the logarithm on the both sides, it is equivalent to
$$
\lambda\ln A+(1-\lambda)\ln B\le\ln(\lambda A+(1-\lambda)B)
$$
which is the same as the definition of the logarithm being a concave function.