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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 for all $x$. For all $x\geq 0$, prove that

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

2 Answers 2

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Since 0 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.

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    Thank you very much @Sivaram.You are absolutely right.2011-12-01
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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 $x>0$.