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证明 $\dps{\int_0^\frac{\pi}{2} t\sex{\frac{\sin nt}{\sin t}}^4\rd t<\frac{\pi^2n^2}{4}}$.
证明: 先回忆两个常用不等式:
(1). $|\sin nx|\leq n|\sin x|$, $\forall\ x$. 这可用数学归纳法证明. 当 $n=1$ 时结论自明. 设当 $n=k$ 时结论成立, 则 $$\beex \bea |\sin (k+1)x|&=|\sin kx\cos x+\cos kx\sin x|\\ &\leq |\sin kx|+|\sin x| \leq k|\sin x|+|\sin x| =(k+1)|\sin x|. \eea \eeex$$
(2). $\dps{\sin x>\frac{2}{\pi}x,\ 0<x<\frac{\pi}{2}}$. 几何上这是明显的. 分析上来看, $f(x)=\sin x$ 是凹函数, 而 $$\bex f(x)=f\sex{\sex{1-\frac{2x}{\pi}}\cdot 0+\frac{2x}{\pi}\cdot \frac{\pi}{2}} >\sex{1-\frac{2x}{\pi}}\cdot f(0)+\frac{2x}{\pi}\cdot f\sex{\frac{\pi}{2}}=\frac{2x}{\pi}. \eex$$ 往证题目. $$\beex \bea \int_0^\frac{\pi}{2} t\sex{\frac{\sin nt}{\sin t}}^4\rd t &=\int_0^\frac{\pi}{2n} +\int_\frac{\pi}{2n}^\frac{\pi}{2}t\sex{\frac{\sin nt}{\sin t}}^4\rd t\\ &\leq \int_0^\frac{\pi}{2n} t\sex{\frac{n\sin t}{\sin t}}^4\rd t +\int_\frac{\pi}{2n}^\frac{\pi}{2} t\sex{\frac{1}{\frac{2t}{\pi}}}^4\rd t\\ &=n^4\frac{t^2}{2}|_0^\frac{\pi}{2n} +\sex{\frac{\pi}{2}}^4\int_\frac{\pi}{2n}^\frac{\pi}{2} \frac{1}{t^3}\rd t\\ &=\frac{\pi^2n^2}{8} +\sex{\frac{\pi}{2}}^4\sex{-\frac{1}{2}t^{-2}}|_{\frac{\pi}{2n}}^\frac{\pi}{2}\\ &<\frac{\pi^2n^2}{8}+\sex{\frac{\pi}{2}}^4 \frac{1}{2} \sex{\frac{2n}{\pi}}^4\\ &=\frac{\pi^2n^2}{4}. \eea \eeex$$
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原文地址：http://www.cnblogs.com/zhangzujin/p/4528950.html
