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In my research the following problem came up, but I don't think that it is interesting enough for mathoverflow.

Let $f(x) := \tan(\frac{\pi}{4} x) + (1 - x) \sin(\frac{\pi}{2} x) - x$ for $x \in (0,1)$. Notice that $f(0)=f(1)=0$. In addition, numerics show that for $x \in (0,1)$, $f(x)>0$. I would like to prove this statement rigorously.

So far I tried solving for $x$ such that the derivative vanish (including solving implicitly and substituting back into the equation), convexity arguments, Taylor expansion and algebraic manipulations. I have not tried to do any kind of rigorous numerics and would really prefer not to go into that direction.

Thanks in advance for anything!

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  • $\begingroup$ @JackD'Aurizio Thanks for the comment! It meant the latter (I fixed the question to resolve the ambiguity now) $\endgroup$ – Shwouchk Dec 12 '16 at 17:24
  • $\begingroup$ All right. Then, by exploiting convexity/concavity, the given inequality boils down to two simple polynomial inequalities. $\endgroup$ – Jack D'Aurizio Dec 12 '16 at 17:31
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On the interval $(0,1)$ we have $\sin\frac{\pi x}{2}> x$ by the concavity of the sine function and $$\tan\frac{\pi x}{4} > \max\left(\frac{\pi x}{4},1+\frac{\pi}{2}(x-1)\right) $$ by the convexity of the tangent function. It follows that over $\left(0,\frac{2\pi-4}{\pi}\right]$ we have: $$ \tan\frac{\pi x}{4}+(1-x)\sin\frac{\pi x}{2}-x > x\left(\frac{\pi}{4}-x\right)>0$$ and over $\left[\frac{2\pi-4}{\pi},1\right)$ we have: $$ \tan\frac{\pi x}{4}+(1-x)\sin\frac{\pi x}{2}-x > (1-x)\left(x+\frac{2-\pi}{2}\right)>0.$$

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  • $\begingroup$ Amazing! Thanks! $\endgroup$ – Shwouchk Dec 12 '16 at 17:41
  • $\begingroup$ (please ignore the deleted comment) $\endgroup$ – Shwouchk Dec 12 '16 at 19:00

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