# If $f'' \ge 0$, for $a, b \in (0, 1)$, prove that $f(ab)\le af(b)$

I tried to use the definition of the concave function, but I failed to transform it into the inequality that has to be proved. The original question is:

If $$f'' \ge 0, f(0)=0$$, and $$f$$ is differentiable, $$\forall a, b \in (0, 1)$$, prove that $$f(ab)\le a f(b)$$

I thought that if $$f$$ is concave up, the following satisfies: $$\frac{f(b)-f(0)}{b-0}\le\frac{f(b-h)-f(b)}{-h}$$ where $$0\le h\le b$$. I still don't know how this satisfies mathematically. When we change the inequality by multiplying $$-h$$: $$f(b-h)\le f(b)-\frac{h}{b}f(b)=\left(1-\frac{h}{b}\right)f(b)$$ When we put $$h=(1-a)b$$, $$f(ab)\le a f(b)$$

Is this the right procedure? Furthermore, I want to know more precise proof. Could you please give me some key points to this question? Thanks for your advice.

Since $$f''$$ is nonnegative, $$f$$ is a convex function. So we can use Jensen's inequality:
For any $$x,y$$ and for any $$t\in[0,1]$$, we have $$f(tx+(1-t)y)\le tf(x)+(1-t)f(y).$$
To get the desired inequality, take $$t=a$$, $$x=b$$ and $$y=0$$.