Is the function $\frac1x\log\sum\exp\left(c_i x^2\right)$ convex for every nonnegative $c_i$s? While reading a machine learning paper, I came across the following statement:

The function $\dfrac{f(x)}{x}$ is convex, where
  $$f(x) = \log\left(\sum_{i = 1}^m \exp\left(c_i x^2\right)\right),$$
  with $c_1, \dots, c_m \geq 0$ and $x>0$.

I know that in general, the log-of-sum-of-exponentials is convex, but why does it remain convex when it is multiplied by $\dfrac{1}{x}$?

Let me assume that $x \geq 0$ as otherwise the function is clearly not convex.
The perspective of the log-sum-exp function is convex, so
$$g(x,y) = \begin{cases} y \log\left(\sum\exp\left(c_i\frac{x}{y}\right)\right) & \text{if } y \geq \frac{1}{x}, x \geq 0 \\
\infty &\text{otherwise}\end{cases}$$
is convex too. The function you are interested in is the partial minimization of $g$:
$h(x) = \inf_y \{ g(x,y) \}$, which is convex (as partial minimizations of convex functions are convex).
