Putnam 2025, Problem A-2

Find the largest real number \( a \) and the smallest real number \( b \) such that
\[
a x (\pi - x) \leq \sin x \leq b x (\pi - x)
\]
for all \( x \) in the interval \( [0, \pi] \).

Solution:
The answer is \( a = 1/\pi \) and \( b = 4/\pi^2 \).

**First solution.** It suffices to restrict to \( x \in [0, \pi/2] \) rather than \( x \in [0, \pi] \) since the given inequalities are unchanged if we replace \( x \) by \( \pi - x \). Define the function
\[
f(x) = \frac{\sin x}{x(\pi - x)},
\]
and note that \( f \) is differentiable on \( (0, \pi/2] \) and can be extended continuously to \( x = 0 \) by setting \( f(0) = 1/\pi \). In particular, if \( a \) and \( b \) satisfy the desired condition, then \( a \leq 1/\pi \) and \( b \geq f(\pi/2) = 4/\pi^2 \); it thus suffices to check that \( a = 1/\pi \) and \( b = 4/\pi^2 \) satisfy the condition. We compute that \( f'(x) = \frac{g(x)}{x^2(\pi - x)^2} \) for
\[
g(x) = x(\pi - x)\cos x - (\pi - 2x)\sin x.
\]
We claim that \( g(x) > 0 \) for all \( x \in (0, \pi/2) \). Indeed, note that \( g(0) = g(\pi/2) = 0 \) and
\[
g'(x) = (x^2 - \pi x + 2)(\sin x).
\]
Noting that the quadratic \( x^2 - \pi x + 2 \) is positive at \( x = 0 \) and negative at \( x = \pi/2 \), we see that it has a root \( r \in (0, \pi/2) \), is positive before that root, and is negative after that root. That is, \( g'(x) > 0 \) for \( x \in (0, r) \) and \( g'(x) < 0 \) for \( x \in (r, \pi/2) \). If some \( x_0 \in (0, \pi/2) \) satisfied \( g(x_0) \leq 0 \), then we would have both \( x_0 > r \) since \( g \) is increasing in \( (0, r) \) and \( g(0) = 0 \), and \( x_0 < r \) since \( g \) is decreasing in \( (r, \pi/2) \) and \( g(\pi/2) = 0 \), contradiction. This proves the claim. Since \( f'(x) > 0 \) on \( (0, \pi/2) \), it follows that \( f(x) \) is strictly increasing on \( [0, \pi/2] \) and hence achieves a minimum at \( f(0) = 1/\pi \) and a maximum at \( f(\pi/2) = 4/\pi^2 \).

**Second solution.** As in the first solution, we reduce to checking that \( a = 1/\pi \) and \( b = 4/\pi^2 \) satisfy the given inequality for \( x \in [0, \pi/2] \). For the lower bound, we use the fact that for \( x \in [0, \pi/2] \),
\[
\sin x \geq x - \frac{1}{6}x^3
\]
(by either Taylor’s theorem with remainder or the alternating series test) and the fact that \( \pi^2 < 12 \) to obtain
\[
\sin x - \frac{1}{\pi}x(\pi - x) \geq x - \frac{1}{6}x^3 - \frac{1}{\pi}x(\pi - x) = x^2\left( \frac{1}{\pi} - \frac{x}{6} \right) \geq x^2\left( \frac{1}{\pi} - \frac{\pi}{12} \right) \geq 0.
\]
For the upper bound, set
\[
f(x) = \frac{4}{\pi}x(\pi - x) - \sin x,
\]
so that \( f(0) = f(\pi/2) = 0 \), then compute that
\[
f'(x) = \frac{4}{\pi} - \frac{8}{\pi}x - \cos x, \quad f''(x) = -\frac{8}{\pi^2} + \sin x.
\]
Since \( \pi^2 > 8 \), \( f''(x) \) has exactly one root in \( [0, \pi/2] \) and \( f''(\pi/2) > 0 \). Since \( f'(0) > 0 \), \( f'(\pi/2) = 0 \), and \( f''(\pi/2) > 0 \), by Rolle’s theorem \( f' \) has exactly one root \( r \) in \( [0, \pi/2) \), at which \( f'' \) is negative. Consequently, the absolute minimum of \( f \) occurs as the endpoints, proving the claim.

**Third solution.** (by Catalan Zara) As in the first solution, we reduce to checking that \( a = 1/\pi \) and \( b = 4/\pi^2 \) satisfy the given inequality for \( x \in [0, \pi/2] \). For the lower bound, note that \( \cos t \) is concave on \( [0, \pi/2] \), so \( \cos t \geq 1 - (2/\pi)t \) and hence
\[
\sin x = \int_0^x \cos t \, dt \geq \int_0^x \left( 1 - \frac{2}{\pi}t \right) dt = \frac{1}{\pi}x(\pi - x).
\]
For the upper bound, note that \( \sin t \) is concave on \( [0, \pi/4] \), so \( \sin t \geq (2\sqrt{2}/\pi)t \) and so
\[
\sin x = \cos\left( \frac{\pi}{2} - x \right) = 1 - 2\sin^2\left( \frac{\pi}{4} - \frac{x}{2} \right) \leq 1 - \frac{16}{\pi^2}\left( \frac{\pi}{4} - \frac{x}{2} \right)^2 = \frac{4}{\pi^2}x(\pi - x).
\]