Question

For $m>0$,
$0 < n\leqslant m+1$ ($m,n\in \mathbb{Z} $) , and $0 < a < 1$ , prove that $$2^{n}\cdot \left( a^{n}\cos ^{2m}\dfrac {\pi a} {2}+\left( 1-a\right) ^{n}\sin ^{2m}\dfrac {\pi a} {2}\right) \leqslant\cos ^{2m}\dfrac {\pi a} {2}+\sin ^{2m}\dfrac {\pi a} {2}. $$

Answer 1

Note that both sides are equal at $a=1/2$, and symmetric under $a \to 1-a$. So let's assume $0 < a < 1/2$. Then the inequality says $$\tan^{2m}(\pi a/2) \le \dfrac{1 - 2^n a^n}{2^n (1-a)^n - 1}$$ Now in fact, since $\tan$ is convex on $[0,\pi/4]$ we have $\tan^{2m}(\pi a/2) \le (2a)^{2m}$, and it suffices to prove that $$ 2^{2n-1} a^{2n-1} \le \dfrac{1 - 2^n a^n}{2^n (1-a)^n - 1}$$ i.e. that $2^{3n-1}a^{2n-1} (1-a)^n \le 1 - 2^n a^n + 2^{2n-1} a^{2n-1}$. In fact, $a(1-a) \le 1/4$, so $2^{3n-1} a^{2n-1}(1-a)^n \le 2^{n-1} a^{n-1}$, and $(1-2^n a^n)(1-2^{n-1}a^{n-1}) \ge 0$ so $2^{n-1} a^{n-1} \le 1 - 2^n a^n + 2^{2n-1} a^{2n-1}$.