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