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

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 $a=\pi^2<10$.
A: 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.
