How show that $\dfrac{a^3}{3}\ge\int_{0}^{a}|F(x)-x|^2dx$ Let $F(x)$ be nonnegative and integrable on $[0,a]$ and such that
$$\left(\int_{0}^{t}F(x)dx\right)^2\ge\int_{0}^{t}F^3(x)dx$$
for every $t$ in $[0,a]$,prove or disprove the conjecture:
$$\dfrac{a^3}{3}\ge\int_{0}^{a}|F(x)-x|^2dx$$
This Problem from SIAM problem 78-18,I consider some time,but I failure it.
 A: For simplicity let us write $F = F(x)$ and $G_t = \int_0^t F dx$.  
Given $\displaystyle {G_t}^2 \ge \int_0^t F^3 dx$,  we need to show $\displaystyle \dfrac{a^3}{3} \ge \int_0^a \mid F - x \mid^2 dx \tag{1}$.
Now, RHS $= \displaystyle  \int_0^a \left( F - x \right)^2 dx = \int_0^a F^2 dx - \int_0^a 2x F dx + \frac{a^3}{3}$.
So we need to show $$\displaystyle \color{blue}{2\int_0^a x F dx} \ge \color{red}{\int_0^a  F^2 dx} \tag{2}$$
By Hölder's inequality, we have:  $$G_t = \int_0^t F dx \le \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot \left( \int_0^t 1 dx \right)^{\frac{2}{3}} =   \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot  t^{\frac{2}{3}} \le {G_t}^{\frac{2}{3}}\cdot t^{\frac{2}{3}} \quad \implies \sqrt{G_t} \le t $$  
So we have: $\displaystyle \int_0^t x \cdot F dx \ge  \int_0^t \sqrt{G_x}\cdot F dx = \frac{2}{3} {G_t}^{\frac{3}{2}}$ and $\color{blue}{\displaystyle 2\int_0^a x F dx \ge \frac{4}{3} {G_a}^{\frac{3}{2}}}$
Also by Hölder, we have:
$$\int_0^t F^2 dx = \int_0^t F^{\frac{3}{2}}\cdot F^{\frac{1}{2}} dx \le \left( \int_0^t F^3 dx \right)^{\frac{1}{2}} \cdot \left( \int_0^t F dx \right)^{\frac{1}{2}} \le G_t \sqrt{G_t} =  {G_t}^{\frac{3}{2}} $$ and $\color{red}{\displaystyle \int_0^a F^2 dx \le {G_a}^{\frac{3}{2}}} $.  Thus (2) holds, and hence (1).
