Prove or disprove the conjecture: $\frac{a^3}{12}\ge\int_{0}^{a}|F(x)-\frac{1}{2}x|^2dx$

Here is a start, mimicking Macavity method. I prove that for all $t\in[0,a]$, $G_t\le \frac 12t^2$.

Proof: Note $k=\max_t\frac{G_t}{t^2}$. For all $t\in[0,a]$, we have $$\int_0^t x^2F(x)\,\mathrm dx\le\left(\int_0^tx^3\,\mathrm dx\right)^{2/3}\left(\int_0^tF^3(x)\,\mathrm dx\right)^{1/3}\le \frac 1{4^{2/3}}t^{8/3}G_t^{2/3}$$ but by definition of $k$ we also have $$\int_0^t x^2F(x)\,\mathrm dx\ge\int_0^t\frac{G_x}kF(x)\,\mathrm dx=\frac{1}{2k}G_t^2$$ Therefore we have $$G_t^{4/3}\le \frac{k\,t^{8/3}}{2^{1/3}}\iff G_t\le\frac{k^{3/4}t^2}{2^{1/4}}$$

Since this inequality holds for all $t\in[0,a]$ we have $$\frac{k^{3/4}}{2^{1/4}}\ge k\iff k\le \frac 12$$

This is not enough to prove your objective: we get $$\int_0^axF(x)\,\mathrm dx\ge \int_0^a\sqrt{2G_x}F(x)\,\mathrm dx=\frac{2\sqrt 2}3G_t^{3/2}$$ and this inequality is optimal for $F(x)=x$, so we have to improve the upper bound on $\displaystyle\int_0^a F^2(x)\,\mathrm dx$.

I wonder if we can reduce the case to nondecreasing $F$, with $A_y=\{x\in[0,a]:F(x)\le y\}$ and $\tilde{F}(x)=\sup\{y:\mu(A_y)\le x\}$, so that $\int_0^a F^{\alpha}(x)\,\mathrm dx=\int_0^a \tilde{F}^{\alpha}(x)\,\mathrm dx$.