Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question

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.

share|cite|improve this answer
Paul, I completely understand the mathematics here. Can one think of having some geometrical/physical meaning for coming up with such a proof? Or is it just from a trial basis? – Ashok Dec 1 '11 at 7:17
@Ashok: This is a nice typical way of proving such functional inequality. As done here, to prove a functional inequality is to first check if the function is monotone and accordingly check if the inequality is satisfied at the desired end points. In essence it is checking whether the inequality is satisfied for the desired extremum of the function. If so, then the inequality is satisfied for the entire function. – user17762 Dec 1 '11 at 7:23
Thank you very much @Sivaram.You are absolutely right. – Ashok Dec 1 '11 at 8:43

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 $x>0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.