Why if $f$ is convex and $ a \ge 0 $ the the function $g(x)=f(x+a)-f(x)$ is increasing.
[Attempt] I was thinking about the following property of convex functions: If $x\lt y \lt z$ then $$ \frac{f(x)-f(y)}{x-y} \le \frac{f(y)-f(z)}{y-z}$$ but I don't know how to get to $x+a$ and $y+a$.
Take $x\le y$. Then $y$ is between $x$ and $y+a$, i.e., $$ y = \lambda x + (1-\lambda) (y+a) $$ for some $\lambda \in (0,1)$. Rearranging yields $\lambda y = \lambda x + (1-\lambda)a$ and $$ x+a = \lambda ( y+a) + (1-\lambda) x, $$ so that $x+a$ is a convex combination of $x$ and $y+a$, with the same coefficients as $y$ is a convex combination of $x$ and $y+a$.
By convexity, we get $$ f(y) \le \lambda f(x) + (1-\lambda) f(y+a) $$ and $$ f(x+a) \le \lambda f(y+a) + (1-\lambda) f(x). $$ Adding both inequalities results in $$ f(x+a) + f(y) \le f(y+a) + f(x)$, $$ or equivalently, $$ f(x+a) - f(x) \le f(y+a) - f(y), $$ and $x\mapsto f(x+a)-f(x)$ is monotonically increasing.
Convexity of $f$ implies $f'$ nondecreasing as well as
$$x\mapsto f(x+a)-f(x)=\int_0^af'(t+x)dt.$$
