If $f''\ge 0$, prove that $f(x+f'(x)) \ge f(x)$

Question

Question:

$f$ 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.

Answer 1

$\int_x^{x+f'(x)} f'(t) dt \geq \int_x^{x+f'(x)} f'(x) dt=(f'(x))^{2}$ if $f'(x) \geq 0$. [I have used the fact that $f'$ is increasing]. Hence $f(x+f'(x))-f(x) \geq (f'(x))^{2}\geq 0$. A similar argument works when $f'(x) <0$.

Answer 2

Hint: If $f$ is twice continuously differentiable then we have $f(x + f'(x)) = f(x) + {f'(x)}^2 + \frac{1}{2} {f'(x)}^2 f''(c) ,$ for some $c$.

Answer 3

Attempt:

MVT:

If $f'(x_0)= 0$, the inequality is obvious.

1)Assume $f'(x_0)>0$; then $f'(x) \ge f'(x_0)>0$ for $x\ge x_0$, since $f'' \ge 0$.

Let $x \ge x_0$:

$\dfrac{f(x+f'(x))-f(x)}{f'(x)}=f'(t)$, $t \in (x,x+f'(x))$.

$f(x+f'(x))-f(x)=$

$f'(x)f'(t)>0;$

Then $f(x+f'(x))\gt f(x)$.

2) Assume $f'(x_1)<0:$

$f'(x) \le f'(x_1) <0$, for $x \le x_1$.

Let $x \le x_1$:

$\dfrac{f(x+f'(x))-f(x)}{f'(x)} =f'(s)<0$, $s \in (x+f'(x),x)$.

Then $f(x+f'(x))\ge f(x)$.

3) interval $ I:= [x_1,x_0]$:

Since $f'$ increasing, there is a zero in $I$.

For $f'(x)<0$, argument $2$;

For $f'(x)>0$, argument $1$.