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

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.

Let $\int\limits_0^1f(x)dx=t$.

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