Show that $\int_0^1 f^3(x) dx + \frac{4}{27} \ge \left( \int_0^1 f(x) dx \right)^2$, where $f',f'' >0$

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.

• +1 Interesting question, OP showed reasonable work trying to solve problem, graphics were legible, and OP provided backround re WhatsApp preparation group. Minor editing suggestion: Re formatting math at mathSE, please see math.stackexchange.com/help/notation. Commented Sep 26, 2020 at 5:55
• There are several good reasons to post questions as text, not as an image. See for example math.meta.stackexchange.com/a/11697/42969. Commented Sep 26, 2020 at 7:02
• @MartinR I absolutely agree with you, out of hand. In this case, I tried to shape my criticism with a light touch, because the graphics were at least legible, and with respect to the OP's showing work + background, his query was (in my opinion) better than 90% of the queries posted on mathSE. Commented Sep 26, 2020 at 7:06

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.
• Just mentioning that the inequality is sharp: Equality holds for the constant function $f(x) = 2/3$. Commented Sep 26, 2020 at 7:17
Let $$\int\limits_0^1f(x)dx=t$$.
Thus, by Holder and AM-GM we obtain: $$\int\limits_0^1f(x)^3dx+\frac{4}{27}=\left(\int\limits_0^11dx\right)^2\int\limits_0^1f(x)^3dx+\frac{4}{27}\geq\left(\int\limits_0^1f(x)dx\right)^3+\frac{4}{27}=$$ $$=t^3+\frac{4}{27}=2\cdot\frac{t^3}{2}+\frac{4}{27}\geq3\sqrt[3]{\left(\frac{t^3}{2}\right)^2\cdot\frac{4}{27}}=t^2=\left(\int\limits_0^1f(x)dx\right)^2.$$