<s>[INST] 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 $?
 [/INST] 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.
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