TITLE: Show that $\int_0^1 f^3(x) dx + \frac{4}{27} \ge \left( \int_0^1 f(x) dx \right)^2$, where $f',f'' >0$
QUESTION [5 upvotes]: Let $f :[0,1] \to [0,\infty)$, $f$ is twice differentiable, $f'(x) >0$, $f''(x) >0$ for all $x\in [0,1]$. Prove that
$$ \int_0^1 f^3(x) dx + \frac{4}{27} \ge \left( \int_0^1 f(x) dx \right)^2.$$
Here $f^3(x)$ power means $f(x)$ raised to power $3$.

I got this problem from our WhatsApp preparation group for math exams from my friend who tried this without success.
My ideas : I had very less ideas about the problem but I did try to use Mean Value theorems without success. Observe that $4$ is a square which is somehow related to the RHS portion of the inequality and $27$ is cube which is somehow might be related to LHS first term. I also observed the fact $f$ is a convex function.
Any help in this problem will be appreciated.

REPLY [3 votes]: Let $\int\limits_0^1f(x)dx=t$.
Thus, by Holder and AM-GM we obtain:
$$\int\limits_0^1f(x)^3dx+\frac{4}{27}=\left(\int\limits_0^11dx\right)^2\int\limits_0^1f(x)^3dx+\frac{4}{27}\geq\left(\int\limits_0^1f(x)dx\right)^3+\frac{4}{27}=$$
$$=t^3+\frac{4}{27}=2\cdot\frac{t^3}{2}+\frac{4}{27}\geq3\sqrt[3]{\left(\frac{t^3}{2}\right)^2\cdot\frac{4}{27}}=t^2=\left(\int\limits_0^1f(x)dx\right)^2.$$