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.
Best Answer
By Holder inequality, $$\left(\int_0^1 f(x) dx\right)^2 \le \left(\int_0^1 f^3(x) dx\right)^{2/3}$$ then by Young's
\begin{align}\left(\int_0^1 f^3(x) dx\right)^{2/3}&= \left(\frac 32 \int_0^1 f^3(x) dx\right)^{2/3}\left( \frac 23\right)^{2/3}\\ &\le \int_0^1 f^3(x) dx + (2/3)^2/3 = \int_0^1 f^3(x) dx + \frac{4}{27} \end{align}
The conditions on $f', f''$ are not used.