muliplication of two convex and inceasing functions is convex Suppose that we have two functions $f$ and $g$. if $f>0 \: and \: g>0$ are both increasing and convex functions, how to prove that $f(x)g(x)$ is also convex?

I think the easiest way would be to show that the directional derivative
$$(fg)'_+(x) = \lim_{h\to 0^+}\frac{(fg)(x + h) - (fg)(x)}{h}$$
is increasing. We have,
\begin{align*}
\frac{(fg)(x + h) - (fg)(x)}{h} &= \frac{(fg)(x + h) - f(x + h)g(x) + f(x + h)g(x) - (fg)(x)}{h} \\
&= f(x + h)\frac{g(x + h) - g(x)}{h} + g(x)\frac{f(x + h) - f(x)}{h} \\
&\to f(x) g'_+(x) + g(x)f'_+(x),
\end{align*}
as $h \to 0^+$.
Because $f$ and $g'_+$ are both positive and increasing (note that $g$ is convex), we have $f \cdot g'_+$ is increasing. The same is true for $g \cdot f'_+$, hence for $(fg)'_+$. Thus, $fg$ is convex.
