Proving Inequality

**mohammadfawaz** · June 2nd 2010, 09:53 AM

how can we prove that $\int_0^1(1-x^2)^ndx \leq \frac{\pi}{2\sqrt{n}}$

I think the proof starts by letting $x=cos(u)$ but I didn't reach anything
Thanks

**Opalg** · June 3rd 2010, 04:03 AM

Originally Posted by mohammadfawaz

how can we prove that $\int_0^1(1-x^2)^ndx \leq \frac{\pi}{2\sqrt{n}}$

I think the proof starts by letting $x=cos(u)$ but I didn't reach anything

I don't know where the $\pi/2$ comes from, but there is a fairly simple way to prove a stronger inequality.

Let $I_n = \int_0^1(1-x^2)^n\,dx$ . Integrate by parts to get $I_n = \Bigl[x(1-x^2)^n\Bigr]_0^1 + 2n\int_0^1x^2(1-x^2)^{n-1}dx = 2n(I_{n-1}-I_n)$ . Therefore $I_n = \frac{2n}{2n+1}I_{n-1}$ .

Using that recurrence relation, you can show by induction that $I_n\leqslant \frac1{\sqrt{n+1}}$ . In fact, the base case n=0 is obvious. The inductive step amounts to the inequality $\frac{2n+2}{2n+3}\,\frac1{\sqrt{n+1}}\leqslant \frac1{\sqrt{n+2}}$ , which you can check by squaring both sides.

Since $\frac1{\sqrt{n+1}}$ is clearly less than $\frac\pi{2\sqrt n}$ , that is a better inequality than what was asked for.

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