Integral inequality: $3\left(\int_0^1 g(x)^2 dx\right)^3 \geq \int_0^1 g(x)^8 dx.$

Question

Problem: Suppose that $g:[0,1]\to\mathbb{R}$ is a differentiable function such that $0\leq g'(x)\leq 1$ for each $x\in[0,1]$ and $g(0)=0.$ Prove that $$3\left(\int_0^1 g(x)^2 dx\right)^3 \geq \int_0^1 g(x)^8 dx.$$

I tried with bounds like $g(x)^m\le m\int_0^x g(t)^{m-1}dt$ but that did not work out. Another small observation is that equality holds here for $g(x)=x,$ so we should give bounds that admit equality for this $g.$

Answer 1

Hint: Show that $\forall x \in [0,1]$,

$$3\left(\int_0^X g(x)^2 dx\right)^3 \geq \int_0^X g(x)^8 dx.$$

Differentiate both sides. What do you get?

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Differentiate both sides. What do you get?

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Differentiate both sides. What do you get?

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Now, do it in reverse.

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If you really really want more hand holding:

Here is a more basic version of this problem. Adapt it.