Prove that $\int_0^a |f(x)|^2|f'(x)|dx \le \frac{a^2}{3}\int_0^a |f'(x)|^3dx$

If $$f \in C^1([0,a])$$ and $$f(0) = 0$$, prove that $$\int_0^a |f(x)|^2|f'(x)|dx \le \frac{a^2}{3}\int_0^a |f'(x)|^3dx$$ when does the equality hold?

My attempt: Since $$f(0) = 0$$, we have $$f(x) = \int_0^x f'(t)dt$$ Plugging it into the LHS of the inequality the LHS becomes $$\int_0^a |f(x)|^2|f'(x)|dx = \int_0^a \left|\int_0^x f'(t)dt \right|^2 f'(x)dx$$ Due to the Cauchy-Bunyakovskiy inequality for $$1$$ and $$f'(t)$$, we have \begin{align} \int_0^a \left|\int_0^x f'(t)dt \right|^2 f'(x)dx &\le \int_0^a\left(\int_0^x 1^2 dt\right)\left(\int_0^x [f'(t)]^2dt \right)f'(x)dx \\ &= \int_0^a \int_0^x xf'(x)[f'(t)]^2dt dx \\ &= \int_0^x [f'(t)]^2dt\int_0^axf'(x)dx \\ &= \left(xf(x)|_0^a - \int_0^a f(x)dx \right)\int_0^x [f'(t)]^2dt \\ &= af(a)\int_0^x [f'(t)]^2dt - \int_0^a f(x)dx \int_0^x [f'(t)]^2dt \end{align} I got stuck here. I cannot figure out the best way to observe the result.

Any help is greatly appreciated.

One can proceed similarly as in Let $f \in AC[0,1],f(0)=0$. Show that $\int_0^1 \lvert f(x)f'(x)\rvert\,dx \leq \int_0^1\lvert f'(x)\rvert^2 \, dx$:
We have $$|f(x)| = \left| \int_0^1 f'(t) \, dt \right| \le \int_0^x |f'(t)| \, dt =: F(x) \, ,$$ with equality for all $$x$$ if and only if $$f'$$ has constant sign on $$[0, a]$$.
Then $$F'(x) = |f'(x)|$$ and $$\int_0^a |f(x)|^2|f'(x)|dx \le \int_0^a F(x)^2 F'(x) \, dx = \frac 13 F(a)^3 \, .$$
Then $$F(a)$$ can be estimated Hölder's inequality (with $$p=3/2$$ and $$q=3$$), this gives $$F(a) = \int_0^a 1 \cdot |f'(x)|\, dx \le a^{2/3} \cdot \left( \int_0^a |f'(x)|^3 \, dt \right)^{1/3} \, ,$$ with equality if and only if $$|f'|$$ is constant on $$[0, a]$$.
Remark: In the same way one can prove that $$\int_0^a |f(x)|^p|f'(x)| \, dx \le \frac{a^p}{p+1}\int_0^a |f'(x)|^{p+1} \, dx$$ for any $$p \ge 1$$.