Prove that function is convex How can it be proved that the function $f(x) = \ln \bigl(\sum\limits_{i=1}^{n} e^{x_{i}}\bigr) $ is convex?
 A: This is a useful fact, which implies that the class of log-convex and log-subharmonic functions are closed under addition.
My unimaginative approach is take the second directional derivative, by way of Taylor expansion. Let's fix $x$ and let  $S = \sum_{i=1}^n e^{x_i}$. Fix a nonzero vector $v$ and consider the one variable function 
$$g(t) = f(x+tv) = \ln \left(\sum e^{x_i}e^{tv_i} \right)$$
Expand each exponential: $$e^{tv_i} = 1 + tv_i+ \frac{t^2}{2} v_i^2 +o(t^2) \tag1$$ Also expand the logarithm: $$\ln (S+\delta) = \ln S + \frac1S \delta - \frac{1}{2S^2}\delta^2 +o(\delta^2) \tag2$$
Plugging in $\delta = t\sum v_ie^{x_i}+ \frac{t^2}{2} \sum v_i^2e^{x_i} +o(t^2)$ into (2), we find that the coefficient of $t^2$ is 
$$
\frac1{2S} \sum v_i^2e^{x_i} - \frac{1}{2S^2}\left(\sum v_ie^{x_i}\right)^2 \tag3
$$
To demonstrate convexity, we must show that (3) is nonnegative. This follows from the Cauchy-Schwarz inequality: 
$$\left(\sum v_ie^{x_i}\right)^2 = \left(\sum e^{x_i/2}\cdot v_ie^{x_i/2}\right)^2
\le S \sum v_i^2e^{x_i}  $$
