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

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.

$$\int_x^{x+f'(x)} f'(t) dt \geq \int_x^{x+f'(x)} f'(x) dt=(f'(x))^{2}$$ 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 $$f'(x) <0$$.

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 $$c$$.

• bro why is that a hint , it's good as solved – aryan bansal Feb 8 at 6:50
• Is the hint derived from Taylor Thm? – ToBY Feb 8 at 7:08

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 $$1$$.