TITLE: Strange result about convexity
QUESTION [19 upvotes]: $f \in C^2([0,1])$ with $f''$ convex and $f(0) = f'(0) = f''(0) = 0$.
Is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?

Source: AoPS

REPLY [2 votes]: Here is a more problem solving approach, since this problem comes from AoPS. Instead of working with $f$, work with $g = f''$ which is a convex function.

Note that it is enough to assume $g(1)=0$. Indeed, take $h(x)=f(x)-\frac{x^2}{2} f''(1)$. Then $h'(x) = f'(x)-xf''(1), h''(x) = f''(x)-f''(1)$. Therefore $h''$ is convex is equivalent to $f''$ convex. Moreover
$$ h''(1)+6h(1)-4h'(1) = f''(1)+6f(1)-4f'(1)$$
so the desired inequality for $h$ is equivalent with the one for $f$. Furthermore $h''(1)=0$.

Assuming that $g(1)=0$ and $g$ is convex, we see that $f''=g, f(0)=f'(0)=0$ implies
$$ f'(t) = \int_0^t g(s)ds, f(t)  =\int_0^t \int_0^s g(\tau) d\tau$$
Therefore, it is enough to prove that
$$ \int_0^1 \int_0^t g(s)ds \geq \frac{2}{3} \int_0^1 g(t)dt.$$
Denoting $h(t) = \int_0^t g(s)ds$ we have $h'(0)=h'(1)=0$ and $h'\leq 0$ since $g$ is convex on $[0,1]$ with $g(0)=g(1)=0$. Then of course
$$ \int_0^1 h(t)dt \geq 2/3 h(1)$$
since $h(1)$ is the minimal value of $h$ on $[0,1]$.


Remark: Compared to the previous answers, I used $f''(0)=g(0)=0$. However, I think it is possible to make a trick similar to point 1 above to see that a similar result holds in the general case.