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I am working on proving the below inequality, but I am stuck.

Let $g$ be a differentiable function such that $g(0)=0$ and $0<g'(x)\leq 1$ for all $x$. For all $x\geq 0$, prove that

$$\int_{0}^{x}(g(t))^{3}dt\leq \left (\int_{0}^{x}g(t)dt \right )^{2}$$

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Since $0<g'(x)$ for all $x$, we have $g(x)\geq g(0)=0$. Now let $F(x)=\left (\int_{0}^{x}g(t)dt \right )^{2}-\int_{0}^{x}(g(t))^{3}dt$. Then

$$F'(x)=2g(x)\left (\int_{0}^{x}g(t)dt \right )-g(x)^3=g(x)G(x),$$

where

$$G(x)=2\int_{0}^{x}g(t)dt-g(x)^2.$$

We claim that $G(x)\geq 0$. Assuming the claim, we have $F'(x)\geq 0$ from the above equality, which implies that $F(x)\geq F(0)=0$, which proves the required statement.

To prove the claim, we have

$$G'(x)=2g(x)-2g(x)g'(x),$$

which is nonnegative since $g'(x)\leq 1$ and $g(x)\geq 0$ for all $x$. Therefore,

$G(x)\geq G(0)=0$ as required.

It’s straightforward: The function $g$ is positive for all $x>0$. Therefore $g'(t)\leq 1$ implies

$$2 g(t)g'(t)\leq 2 g(t)\qquad(t>0)\ ,$$

and integrating this with respect to $t$ from $0$ to $y>0$ we get

$$g^2(y)\leq 2\int_0^y g(t)\ dt\qquad(y>0)\ .$$

Multiplying with $g(y)$ again we have

$$g^3(y)\leq 2 g(y)\ \int_0^y g(t)\ dt ={d\over dy}\left(\Bigl(\int_0^y g(t)\ dt\Bigr)^2\right) \qquad(y>0)\ ,$$

and the statement follows by integrating the last inequality with respect to $y$ from $0$ to $x>0$.

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