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Last edited by realnumber 2014-2-17 20:08证得很难看，好歹算是完成了.
1.当$x\in [0,\frac{\pi}{5}]$时，可以证明$w\le \frac{4}{3}x$如下
记$f(x)=w-\frac{4}{3}x,f(0)=0,f'(x)=w'-\frac{4}{3}$,
以下证明$f'(x)\le0 \Leftrightarrow \frac{2\cos{2x}+2\cos{x}}{3\cos{w}}\le \frac{4}{3}$
$\Leftrightarrow {(\cos{2x}+\cos{x})}^2\le 4(1-{(\frac{\sin{2x}+2\sin{x}}{3})}^2)$
$\Leftrightarrow 32\cos^4{x}+28\cos^3{x}-27\cos^2{x}-10\cos{x}\le 31$,显然有$0\le \cos{x} \le 1,32+28-27-10<31$成立.
而$(\pi-2x+w)(x+w)^2\le (\pi-2x+\frac{4}{3}x)(x+\frac{4}{3}x)^2\le 8x^2(\pi-2x)$
$\Leftrightarrow 334x \le 69\pi$,对于$x\in [0,\frac{\pi}{5}]$成立.
2.当$x\in [\frac{\pi}{5},\frac{\pi}{3}]$时,
按如下处理$w\le g(x)=-\frac{9}{2\pi}(x-\frac{\pi}{3})^2+\frac{\pi}{3}$----(1)
$(\pi-2x+g(x))(x+g(x))^2\le 8x^2(\pi-2x)$------(2)
若(1)(2)都成立，则1楼要证明的不等式成立.
先证明(2),记$h(x)=8x^2(\pi-2x)-(\pi-2x+g(x))(x+g(x))^2$
令$\frac{\pi}{3}-x=\pi t,x\in [\frac{\pi}{5},\frac{\pi}{3}],t\in [0,\frac{2}{15}]$
\[\frac{8\times 27h(x)}{{\pi}^3}=64(1-3t)^2(6t+1)-(4+12t-27t^2)(4-6t-27t^2)^2=19683t^6-11664t^4+3024t^3\]
(注：多项式化简来自这个地址zh.numberempire.com/simplifyexpression.php ，又本来打算碰到个缺常数项的6次，继续用导数.而化简结果如此出乎意料.)
如此只需要证明$19683t^3-11664t+3024\ge 0,t\in [0,\frac{2}{15}]$,显然有$3024\ge 11664\times \frac{2}{15}$,那么不等式(2)成立.
要证明不等式(1)，
只需要证明$f(x)=\sin{(-\frac{9}{2\pi}(x-\frac{\pi}{3})^2+\frac{\pi}{3})}-\sin{w}\ge 0,x\in [\frac{\pi}{5},\frac{\pi}{3}]$
而$f(\frac{\pi}{3})=0$,
$f'(x)=\cos{(-\frac{9}{2\pi}(x-\frac{\pi}{3})^2+\frac{\pi}{3})}(-\frac{9}{\pi}(x-\frac{\pi}{3}))-\frac{2\cos{2x}+2\cos{x}}{3}\le 0$
此式对$x\in [\frac{\pi}{4},\frac{\pi}{3}]$成立，如下
令$\frac{\pi}{3}-x=t,t\in [0,\frac{\pi}{12}]$,
$\cos{2x}+\cos{x}=(1+2\cos{t})(\sin^2{\frac{t}{2}}+\frac{\sqrt{3}}{2}\sin{t})\ge \frac{\sqrt{3}}{2}(1+2\cos{t})\sin{t}\ge \frac{\sqrt{3}}{2}(1+2\cos{\frac{\pi}{12}})\sin{t}$
$x\in [\frac{\pi}{4},\frac{\pi}{3}]$时，$-\frac{9}{2\pi}(x-\frac{\pi}{3})^2+\frac{\pi}{3}\ge \frac{\pi}{3}-\frac{\pi}{32}$
要证明$f'(x)\le 0$,只需要证明$\cos{(\frac{\pi}{3}-\frac{\pi}{32})}\frac{9}{\pi}t\le \frac{2}{3}\frac{\sqrt{3}}{2}(1+2\cos{\frac{\pi}{12}})\sin{t}$
由函数图象，只需要在$t=\frac{\pi}{12}$时成立即可，经计算器检验成立.
当$x\in [\frac{\pi}{5},\frac{\pi}{4}]$，
$f''(x)=-\frac{9}{\pi}\cos{(-\frac{9}{2\pi}(x-\frac{\pi}{3})^2+\frac{\pi}{3})}-\frac{81}{{\pi}^2}(x-\frac{\pi}{3})^2\sin{(-\frac{9}{2\pi}(x-\frac{\pi}{3})^2+\frac{\pi}{3})}+\frac{4\sin{2x}+2\sin{x}}{3}$
此时$f''(x)\le-\frac{9}{\pi}\cos{(-\frac{9}{2\pi}(\frac{\pi}{4}-\frac{\pi}{3})^2+\frac{\pi}{3})}-\frac{81}{{\pi}^2}(\frac{\pi}{4}-\frac{\pi}{3})^2\sin{(-\frac{9}{2\pi}(\frac{\pi}{5}-\frac{\pi}{3})^2+\frac{\pi}{3})}+\frac{4\sin{2\frac{\pi}{4}}+2\sin{\frac{\pi}{4}}}{3}=-0.2658\le 0$
即$y=f(x),x\in [\frac{\pi}{5},\frac{\pi}{4}]$为上凸函数，又$f(\frac{\pi}{5})\ge 0,f(\frac{\pi}{4})\ge 0$,所以有$f(x)\ge0$.
至此1楼不等式成立. 无敌分类讨论暴力流~~~ |
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Tesla35
Posted 2014-2-8 12:44
