IMC 2014 · Day 2 · P8

Putting $f(0) = 1$ we can assume that the function is analytic in $\mathbb{R}$. Let $g(x) = x^{n+1} \left( f^{(n)}(x) - \frac{1}{n+1} \right)$. Then $g(0) = 0$ and $$\begin{align*} g'(x) &= (n+1) x^n \left( f^{(n)}(x) - \frac{1}{n+1} \right) + x^{n+1} f^{(n+1)}(x) = \\ &= x^n \Bigl( (n+1) f^{(n)}(x) + x f^{(n+1)}(x) - 1 \Bigr) = x^n \bigl( (x f(x))^{(n+1)} - 1 \bigr) = x^n \bigl( \sin^{(n+1)}(x) - 1 \bigr) \le 0. \end{align*}$$ Hence $g(x) \le 0$ for $x > 0$. Taking into account that $g'(x) < 0$ for $0 < x < \frac{\pi}{2}$ we obtain the desired (strict) inequality for $x > 0$.

$$\left( \frac{\sin x}{x} \right)^{(n)} = \frac{d^n}{dx^n} \int_0^1 -\cos(xt)\,dt = \int_0^1 \frac{\partial^n}{\partial x^n} (-\cos(xt))\,dt = \int_0^1 t^n g_n(xt)\,dt$$ where the function $g_n(u)$ can be $\pm \sin u$ or $\pm \cos u$, depending on $n$. We only need that $|g_n| \le 1$ and equality occurs at finitely many points. So, $$\left| \left( \frac{\sin x}{x} \right)^{(n)} \right| \le \int_0^1 t^n \left| g_n(xt) \right| dt < \int_0^1 t^n\,dt = \frac{1}{n+1}.$$