Q:

Test Function Convexity

I'm trying to test if $ f : \mathbb{R}^n\rightarrow \mathbb{R} $ given by :
$$ f(x) = \ln \left(\sum\limits_{i=1}^n e^{x_i}\right) $$ is convex.
I tried computing positivity of the Hessian  $H$ matrix by definition, multiplying by $a \in \mathbb{R}^n $. I got the expression 
$$ a^T H a = \displaystyle\sum_{i=1}^n a_i^2 e^{x_i} -(a_1 + ... + a_n)\frac{a_1 e^{x_1} + ... + a_n e^{x_n}}{e^{x_1} + ... + e^{x_n} } $$
Where $ a_i $ is the $i$-th coordinate of $a$.
Is there a way to check if this expression is higher than zero? I can't see if an averages inequality would help.

A:

Let $\lambda, \mu \in \langle 0,1\rangle$ be such that $\lambda + \mu = 1$ and $x = (x_1, \ldots, x_n), y= (y_1, \ldots, y_n) \in \mathbb{R}^n$. We wish to show that $$f(\lambda x + \mu y) \le \lambda f(x) + \mu f(y)$$
We shall use Hölder's inequality for conjugate exponents $\frac1{\lambda}$ and $\frac{1}\mu$. We have:
\begin{align}
f(\lambda x + \mu y) &= \ln \sum_{i=1}^n e^{\lambda x_i + \mu y_i} \\
&= \ln \sum_{i=1}^n e^{\lambda x_i}e^{\mu y_i}\\
&\le \ln\, \left[\left(\sum_{i=1}^n e^{\lambda x_i\cdot \frac1\lambda}\right)^{\lambda}\left(\sum_{i=1}^n e^{\mu y_i\cdot \frac1\mu}\right)^{\mu}\right]\\
&= \lambda \ln \sum_{i=1}^n e^{x_i} + \mu \ln \sum_{i=1}^n e^{y_i}\\
&= \lambda f(x) + \mu f(y)
\end{align}
Hence, $f$ is convex on $\mathbb{R}^2$.

