# Prove that $\sin (\pi x) \leq 4x (1-x)$ for all non negative $x \leq 1$

I am interested in a method which uses appropriate mean value theorems. I tried Cauchys mean value theorem but failed. Any help would be appreciated.

Please note:I am not looking for an answer which uses derivates, only mean value theorems (of course, MVTs include derivatives, but I want to solve this question as an application of a MVT)

• "not derivaties, only mean value theorems" Well, (most) mean value theorems involve derivatives, no? – leonbloy Apr 20 at 12:42
• @leonbloy I think he means don't use derivatives to find a maximum/minimum to solve the problem – Digitalis Apr 20 at 12:51
• @leonbloy I have edited the question – N.S.JOHN Apr 20 at 13:00
• If a differential approach is forbidden, then what is the definition of the sine ? – Yves Daoust Apr 20 at 13:02
• @YvesDaoust can you tell me why it matters? – N.S.JOHN Apr 20 at 13:05

It is enough to consider $$x\le \frac12$$, because for $$x>\frac12$$ we can use $$\sin \pi x = \sin(\pi(1-x))$$.
Let us substitute $$t = \frac12-x$$. \begin{align} \sin \pi x &= \sin (\frac{\pi}{2} - \pi t) = \cos \pi t = 1 - 2\sin^2\frac{\pi t}{2}\end{align} Since $$\sin y$$ is a concave function on the interval $$y\in[0,\pi/4]$$ (a fact that you can prove using the MVT, if you want), we have $$\sin y \ge \frac{y \sin \frac{\pi}{4} + (\frac{\pi}{4}-y) \sin 0}{\frac{\pi}{4}} = \frac{2\sqrt{2}y}{\pi}$$ Using this inequality we get $$\sin\pi x \le 1 - 2(\sqrt{2}t)^2 = 1-4t^2 = 4x(1-x)$$
\begin{align} \sin(\pi x)\le4x(1-x)\text{ for }x\in[0,1] &\iff\cos(\pi x)\le1-4x^2\text{ for }x\in\left[-\tfrac12,\tfrac12\right]\tag1\\ &\iff\cos(\pi x)\le1-4x^2\text{ for }x\in\left[\,0,\tfrac12\,\right]\tag2 \end{align} Explanation:
$$(1)$$: substitute $$x\mapsto\frac12-x$$
$$(2)$$: both sides of the inequality are even
For $$x\in\left[\,0,\frac12\,\right]$$, consider $$f(x)=4x^2+\cos(\pi x)\tag3$$ $$f(0)=1$$, $$f'(0)=0$$ and $$f''(x)=8-\pi^2\cos(\pi x)\tag4$$ $$f''(x)\le0$$ for $$0\le x\le x_0=\frac1\pi\cos^{-1}\left(\frac8{\pi^2}\right)$$. The Mean Value Theorem, applied once, says that $$f'(x)\le0$$ for $$0\le x\le x_0$$; and applied twice, says that $$f(x)\le1\text{ for }0\le x\le x_0\tag5$$ $$f''(x)\ge0$$ for $$x_0\le x\le\frac12$$; that is, $$f$$ is convex on $$\left[\,x_0,\frac12\,\right]$$. Since $$f(x_0)\le1$$ and $$f\!\left(\frac12\right)=1$$, we know that $$f(x)\le1\text{ for }x_0\le x\le\tfrac12\tag6$$ $$(2)$$ is satisfied by $$(5)$$ and $$(6)$$.