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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)

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  • $\begingroup$ "not derivaties, only mean value theorems" Well, (most) mean value theorems involve derivatives, no? $\endgroup$ – leonbloy Apr 20 at 12:42
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    $\begingroup$ @leonbloy I think he means don't use derivatives to find a maximum/minimum to solve the problem $\endgroup$ – Digitalis Apr 20 at 12:51
  • $\begingroup$ @leonbloy I have edited the question $\endgroup$ – N.S.JOHN Apr 20 at 13:00
  • $\begingroup$ If a differential approach is forbidden, then what is the definition of the sine ? $\endgroup$ – Yves Daoust Apr 20 at 13:02
  • $\begingroup$ @YvesDaoust can you tell me why it matters? $\endgroup$ – N.S.JOHN Apr 20 at 13:05
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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 $$\sin\pi x \le 1 - 2(\sqrt{2}t)^2 = 1-4t^2 = 4x(1-x) $$

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$$ \begin{align} \sin(\pi x)\le4x(1-x)\text{ for }x\in[0,1] &\iff\cos(\pi x)\le1-4x^2\text{ for }x\in\left[-\tfrac12,\tfrac12\right]\tag1\\ &\iff\cos(\pi x)\le1-4x^2\text{ for }x\in\left[\,0,\tfrac12\,\right]\tag2 \end{align} $$ 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 $(6)$.

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