$f \in C^2([0,1])$ with $f''$ convex and $f(0) = f'(0) = f''(0) = 0$.


Is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?




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Source: [AoPS](https://artofproblemsolving.com/community/c7h1570936)



Strange result about convexity
The answer is yes, and it follows immediately from a simple description of the extreme rays of the convex cone of the functions with a convex second derivative.


Indeed, for $x\in(0,1]$, let $f'''(x)$ denote the left derivative of the convex function $f''$ at $x$, so that $f'''$ is non-decreasing on $(0,1]$.


Take any $x\in(0,1]$. Then, taking into account the conditions $f(0)=f'(0)=0$, one has the Taylor-like formula (cf. e.g. [formula (4.12)](https://projecteuclid.org/journals/banach-journal-of-mathematical-analysis/volume-10/issue-4/Convex-cones-of-generalized-multiply-monotone-functions-and-the-dual/10.1215/17358787-3649788.full) or [formula (3.2)](https://projecteuclid.org/journals/annals-of-statistics/volume-22/issue-1/Extremal-Probabilistic-Problems-and-Hotellings-T2-Test-Under-a-Symmetry/10.1214/aos/1176325373.full)):
\begin{equation\*}
 f(x)=\frac{f''(0)}2\,x^2+\frac{f'''(0+)}6\,x^3+\frac16\int\_{(0,1)}df'''(t)(x-t)\_+^3, \tag{1}
\end{equation\*}
where $z\_+:=\max(0,z)$ and the integral is a Lebesgue--Stieltjes one
(see a detailed derivation of this formula at the end of this answer.)


So, $f$ is a "mixture" of the functions $x\mapsto g\_2(x):=x^2$, $x\mapsto g\_3(x):=x^3$, and $x\mapsto f\_t(x):=(x-t)\_+^3$ for $t\in(0,1)$, and the "coefficients" of the functions $f\_t$ in this mixture are nonnegative.


Therefore, it remains to note that (i) $f''(1)+6f(1)=4f'(1)$ if $f=g\_2$ or $f=g\_3$ and (ii) $f''(1)+6f(1)\geq 4f'(1)$ if $f=f\_t$ for some $t\in(0,1)$ (in the latter case, the inequality $f''(1)+6f(1)\ge 4f'(1)$ is the obvious inequality $6z+6z^3\ge12z^2$ with $z:=1-t$). $\qquad\Box$


The condition $f''(0)=0$ was not needed or used here.




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**Details on the Taylor-like formula (1):**


Take any $x\in(0,1]$. Then
\begin{equation\*}
\begin{aligned}
 f''(x)&=f''(0)+\int\_0^x du\,f'''(u) \\ 
 &=f''(0)+\int\_0^x du\,\Big(f'''(0+)+\int\_{(0,u)}df'''(t)\Big) \\ 
 &=f''(0)+f'''(0+)x+\int\_0^x du\,\int\_{(0,u)}df'''(t) \\ 
 &=f''(0)+f'''(0+)x+\int\_{(0,x)}df'''(t)\int\_t^x du\, \\ 
 &=f''(0)+f'''(0+)x+\int\_{(0,x)}df'''(t)(x-t) \\ 
 &=f''(0)+f'''(0+)x+\int\_{(0,1)}df'''(t)(x-t)\_+; 
\end{aligned} 
\end{equation\*}
the fourth equality in the above display is an instance of the Fubini--Tonelli theorem.


Therefore and by a Taylor formula and the conditions $f(0)=f'(0)=0$,
\begin{equation\*}
\begin{aligned}
 f(x)&=\int\_0^x du\,f''(u)(x-u) \\ 
 &=\int\_0^x du\,\Big(f''(0)+f'''(0+)x+\int\_{(0,1)}df'''(t)(x-t)\_+\Big)(x-u) \\ 
 &= \frac{f''(0)}2\,x^2+\frac{f'''(0+)}6\,x^3+\frac16\int\_{(0,1)}df'''(t)(x-t)\_+^3. 
\end{aligned} 
\end{equation\*}
One can also check this Taylor-like formula by taking the integral $\int\_{(0,1)}df'''(t)(x-t)\_+^3$ "by parts" three times -- that is, by applying the Fubini--Tonelli theorem to this integral three times.


