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For what values of $a>0$ is $\left[2^{-2(a-1)}\sin(\pi x)\right]\leq \left[x^{a-1}(1-x)^{a-1}\right]$ for all $0\le x\le 1$?

This is obviously true for $a\leq 1$, where the LHS is always $\leq 2^{-2(a-1)}$, and the RHS is always $\geq 2^{-2(a-1)}$ (because the minimum is at $\frac{1}{2}$). The functions "touch" at $\frac{1}{2}$. I also believe, on the basis of plotting, that it ceases to be true at $a$ just above 2. But I can't find a general way to prove that it is true for all $a$ such that $1<a<2$.

Any ideas?

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Oops, forgot to specify that $0\leq x \leq 1$. –  RandomUser Feb 1 '12 at 20:07
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2 Answers

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When $x\to0$, the LHS is equivalent to $4^{1-a}\pi x$ and the RHS is equivalent to $x^{a-1}$. If $a\gt2$, $x^{a-1}\ll x$ hence the inequality cannot be true on a neighborhood of $0$.

If $a=2$, the inequality holds if and only if $u(x)\geqslant0$ for every $x$ in $(0,1)$, where $$ u(x)=4x(1-x)-\sin(\pi x). $$ Since $u(1-x)=u(x)$, one can, and we will, assume that $x$ is in $(0,1/2)$. Since $u''(x)=-8+\pi^2\sin(\pi x)$ increases from $u''(0)=-8$ to $u''(1/2)=\pi^2-8\gt0$, $u''$ is first negative then positive on the interval $(0,1/2)$, hence $u'$ decreases then increases. Since $u'(x)=4(1-2x)-\pi\cos(\pi x)$, $u'(0)=4-\pi\gt0$ and $u'(1/2)=0$. Thus $u'$ is positive then negative on the interval $(0,1/2)$. Since $u(0)=u(1/2)=0$, $u\geqslant0$ on the interval $(0,1/2)$. This concludes the proof for $a=2$.

For a given $x$ in $(0,1)$, the ratio of the RHS by the LHS is $(4x(1-x))^{a-1}$ times a factor independent on $a$. This is a nonincreasing function of $a$ because $0\leqslant4x(1-x)\leqslant1$. Hence the inequality being true for $a$ implies it is true at every smaller value of $a$.

Finally the inequality holds for every $a\leqslant2$ and it fails for every $a\gt2$.

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For $0<x<1$ and $a>1$, the inequality $2^{-2(a-1)}\sin(\pi x) \le x^{a-1}(1-x)^{a-1}$ is equivalent to $$ a-1 \le \frac{\log(\sin(\pi x))}{\log(4x(1-x))} $$ (get $\sin(\pi x)$ on one side of the inequality and everything else on the other side, then take logs). You can check that the function on the right-hand side has infemum equal to 1 on the interval $(0,1)$, which proves what you want.

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