Integral Inequality and Convexity 
Let $$f(x)=2^{\int_0^\infty\frac{3^{-tx}}{1+3^{-t}}dt}$$Prove $$\forall x,y>0,\;\;2f(x+y)\leq f(2x)+f(2y)$$

In my attempt to solve this problem I came to the conclusion that it suffices to prove $f(x)$ is a convex function. The definition of convexity does not help me since what I want to prove is easier than the definition of convexity in general. I also know the combination of two convex functions is convex.Question From Jalil Hajimir
 A: Perform twice differentiation you get
\begin{align*}
& f''(x)\\
&=(\log 2)^{2}2^{-\int_{0}^{\infty}\frac{3^{-tx}}{1+3^{-t}}dt}\left(\int_{0}^{\infty}\dfrac{(\log 3)3^{-tx}(-t)}{1+3^{-t}}dt\right)^{2}+(\log 2)2^{\int_{0}^{\infty}\frac{3^{-tx}}{1+3^{-t}}dt}\int_{0}^{\infty}\dfrac{3^{-tx}t^{2}}{1+3^{-t}}dt>0,
\end{align*}
so $f$ is convex.
A: That $f(x)$ is convex follows from the following two facts:   
Fact 1: $\mathrm{e}^{g}$ is convex if $g$ is convex.  
Fact 2: If $g(x,y)$ is convex in $x$ for each $y\in S$,
then $h(x) = \int_S g(x,y) dy$ is convex (provided the integral exists). 
Remark: Their proof is easy and thus omitted. Or see page 85 and page 79, [1].
Indeed, let 
$$F(x, t) = \frac{3^{-tx}}{1 + 3^{-t}}.$$
From Fact 1, $F(x, t)$ is convex in $x$ for each $t$, by noting that $3^{-tx} = \mathrm{e}^{-tx\ln 3}$.
From Fact 2, $\int_0^\infty \frac{3^{-tx}}{1 + 3^{-t}} \mathrm{d}t$ is convex.
From Fact 1 again, $2^{\int_0^\infty \frac{3^{-tx}}{1 + 3^{-t}} \mathrm{d}t}$ is convex.
[1] Boyd and Vandenberghe, "Convex optimization". 
http://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf
