prove inequality $\frac{\sin \pi x}{\pi x} \geq \frac{1-x^2}{1+x^2}$ for all R

Question

Let $F(x) = \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} $, is even, so just prove $x > 0$.

When $ x \geq 1 $, $\frac{\sin\pi x}{\pi x} \leq 1$, so $$ \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} \\= -\frac{\sin\pi (x-1)}{\pi(x-1)} \frac{x-1}{x} - \frac{1-x^2}{1+x^2} \\ \geq \frac{1-x}{x} - \frac{1-x^2}{1+x^2} \\ = \frac{(x-1)^2}{x(1+x^2)} \\ \geq 0 $$

So how to prove it when $0 < x <1 $?

Answer 1

Disclaimer: Here is a solution which is not mine. I just slightly adapted it from the very same question, asked as Problem 5642, by R. Redheffer, American Mathematical Monthly 76, (1969), p.422, entitled "A delightful inequality"...

Let us fix $x$ with $0 < x < 1$:

Starting from the classical infinite product for the cardinal sine:

$$\frac{\sin \pi x}{\pi x}=\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2}\right)=(1-x^2)\underbrace{\prod_{k=2}^{\infty}\left(1-\frac{x^2}{k^2}\right)}_{\lim_{n \to \infty} P_n(x)}$$

where $P_n(x)$ is defined by:

$$P_n(x):=\prod_{k=2}^{n}\left(1-\frac{x^2}{k^2}\right),$$

it is enough to prove that $$\text{for any} \ \ n \ge 2, \ \ P_n(x) \ge \dfrac{1}{1+x^2} \ \ \iff \ \ (1+x^2)P_n(x) \ge 1 \ \ \tag{1}$$

Let us remark the evident relation:

$$P_{n+1}(x)=\left(1-\frac{x^2}{(n+1)^2}\right)P_n(x) \tag{2}$$

Actually we get, for all $n \ge 2$:

$$(1+x^2)P_n(x) \ge 1+\frac{x^2}{n}$$

by a simple induction argument based on (2). Therefore, (1) is established.

An interesting article where I have found the reference to this question can be found here.

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Edit: I found the proof by River Li very interesting.

I noticed a possible alternative treatment beginning at the level where a third degree polynomial appears between square brackets.

Setting $a:=\pi^2$ and $Y:=y^2$,

$$\begin{align} &\Big[2\pi^2(2y^2 + 6\pi^2) - (\pi^2 - y^2)(\pi^2 + y^2)^2\Big]\\ \ge & \ a(2Y+6a)-(a-Y)(a+Y)^2\\ = & \ Y^3 + aY^2 - a^2Y + 4aY + 12a^2- a^3\\ = & \ \underbrace{(Y+a/3)^3}_{>0}+4a\underbrace{\Big[Y(1-a/3)+a(3-7a/27)\Big]}_{p(Y)} \end{align}$$

with first degree polynomial $p(Y)>0$ for $Y \in [0,1]$ because

$$p(0)=a(3-7a/27)>0 \ \ \text{and} \ \ p(1)=a(8/3-7a/27)+1>0$$

as a consequence of the fact that $a=\pi^2<10$.

Answer 2

Equivalent problem: Let $0 < y < \pi$. Prove that $\frac{\sin y}{y} \ge \frac{\pi^2 - y^2}{\pi^2 + y^2}$.

It suffices to prove that, for all $y$ in $[0, \pi]$, $$\sin y - y \frac{\pi^2 - y^2}{\pi^2 + y^2} \ge 0.$$ Denote LHS by $g(y)$. We have the following result. The proof is given at the end.

Fact 1: If $y\in (0, \pi)$ satisfying $g'(y) = 0$, then $g(y) \ge 0$.

Now, recall that a continuous function on a bounded and closed interval achieves its minimum at either endpoints or the interior points with zero derivative. Thus, by Fact 1, noting also that $g(0) = g(\pi) = 0$, we have $g(y) \ge 0$ on $[0, \pi]$. We are done.

$\phantom{2}$

Proof of Fact 1: From $g'(y) = 0$, we have $\cos y = \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}$. We have, \begin{align} &(\sin y)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\ =\ & 1 - (\cos y)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\ =\ & 1 - \left(\frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\ =\ & \left(1 + \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right)\left(1 - \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right) - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\ =\ & \frac{2\pi^2(\pi^2 - y^2)}{(\pi^2 + y^2)^2}\cdot\frac{2y^4 + 6\pi^2y^2}{(\pi^2 + y^2)^2} - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\ =\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4} \Big[2\pi^2(2y^2 + 6\pi^2) - (\pi^2 - y^2)(\pi^2 + y^2)^2\Big]\\ =\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[y^6+ \pi^2y^4+(-\pi^4+ 4\pi^2)y^2 + ( - \pi^6 + 12\pi^4)\Big]\\ \ge\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[\pi^2y^4+(-\pi^4+ 4\pi^2)y^2 + ( - \pi^6 + 12\pi^4)\Big]\\ \ge\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[ 2 \sqrt{\pi^2y^4 \cdot (- \pi^6 + 12\pi^4)} + (-\pi^4+ 4\pi^2)y^2\Big]\tag{1}\\ =\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[ 2\pi \sqrt{- \pi^2 + 12} + (-\pi^2+ 4)\Big]\pi^2 y^2\\ \ge\ & 0 \end{align} where we have used AM-GM inequality in (1).

We are done.