$f$ and $f'$ are differentiable, and $f''\ge 0$. Then, prove that $\forall x \in \mathbb R$, $f(x+f'(x))\ge f(x)$.
Since $f''\ge 0$, I'd like to apply Jensen's theorem, which is: $$f(tx_1 + (1-t)x_2) \le tf(x_1) + (1-t)f(x_2) $$
However, it was hard to determine the value of $x_1$ and $x_2$. Another way came up to my mind was to set the new function $$g(x)=f(x+f'(x))-f(x)$$ and prove that $g(x)\ge 0$ by using $g'(x)$. Unfortunately, when we calculate the derivation of $g(x)$ as following: $$ g'(x)= f''(x)f'(x+f'(x))-f'(x)$$ eventually, there was nothing I can find.
Could you give some key points to this proof? Thanks for your advice.