若0 < x,y < \frac {\pi}{2}求证：\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}
乍一看，没什么思路，那我们就先让他变形：\frac {\frac{x}{\sin x} + \frac{y}{\sin y}}{2} < \frac{1}{\cos \frac{x+y}{2}}
依据就不用多说了，利用\csc x = \frac{1}{\sin x}
那么我们要解决的就是这个\frac{x}{\sin x} 和 \cos \frac{x+y}{2}的关系
下面就让我们高高兴兴的来一个简简单单的变形，其中还用到另一结论：x≤tan x\
\frac{x}{\sin x} = \frac{x}{2\sin \frac{x}{2} \cos \frac{x}{2}} = \frac{x}{4\sin \frac{x}{4}\cos\frac{x}{4} \cos \frac{x}{2}} = \frac{\frac{x}{4}}{\tan \frac{x}{4} \cos^2\frac{x}{4}\cos\frac{x}{2}} \le \frac{1}{\cos^2\frac{x}{4}\cos\frac{x}{2}}
同样的，我们对右边变形，使用和差化积公式
\cos \frac{x+y}{2} = \cos \frac{x}{2} \cos \frac{y}{2} - \sin\frac{x}{2}\sin \frac{y}{2} \le \cos \frac{x}{2} \cos \frac{y}{2}.
也就转化成证明以下两个式子：
\frac{1}{\cos^2\frac{x}{4}\cos\frac{x}{2}} + \frac{1}{\cos^2\frac{y}{4}\cos\frac{y}{2}} \le \frac{2}{\cos \frac{x}{2} \cos \frac{y}{2}}
\frac{2}{(1 + \cos\frac{x}{2})\cos\frac{x}{2}} + \frac{2}{(1 + \cos \frac{y}{2})\cos\frac{y}{2}} \le \frac{2}{\cos \frac{x}{2} \cos \frac{y}{2}}
那么左边也就简简单单地变形，从而得到结论。
\mathrm{LHS} \le \frac{2}{(\cos \frac{y}{2} + \cos\frac{x}{2})\cos\frac{x}{2}} + \frac{2}{(\cos\frac{x}{2} + \cos \frac{y}{2})\cos\frac{y}{2}} = \mathrm{RHS}.
Q.E.D.
第二种方式，我们利用函数的增减性来解决这个问题(解答来自于StackExchange)
下面纯英文解答，请注意
Lemma 1. f(x)=\frac{x}{\sin x} is a positive, increasing and convex function on I=\left(0,\frac{\pi}{2}\right).
Let us assume that \mu = \frac{x+y}{2}\in I is fixed and x\leq y. Let us set \delta=\frac{y-x}{2}. By Lemma 1,
\sup_{\substack{0\leq \delta < \min\left(\mu,\frac{\pi}{2}-\mu\right)\ }}\left(\frac{\mu-\delta}{\sin(\mu-\delta)}+\frac{\mu+\delta}{\sin(\mu+\delta)}\right) is achieved at the right endpoint of the range for \delta. If \mu\leq\frac{\pi}{4} such supremum equals 1+\frac{2\mu}{\sin(2\mu)}.
If \mu\geq\frac{\pi}{4} such supremum equals \frac{\pi}{2}+\frac{2\mu-\pi/2}{\sin(2\mu-\pi/2)}. The inequality
\frac{\pi}{2}+\frac{2\mu-\pi/2}{\sin(2\mu-\pi/2)}\leq \frac{2}{\cos\mu}
over the interval \left[\frac{\pi}{4},\frac{\pi}{2}\right) is very loose, hence the problem boils down to showing that
1+\frac{2\mu}{\sin(2\mu)} \leq \frac{2}{\cos\mu}
holds over the interval \left(0,\frac{\pi}{4}\right). By multiplying both sides by \sin\mu\cos\mu we get that the inequality is equivalent to
\sin\mu\cos\mu + \mu \leq 2\sin\mu\tag{E}
which follows from the termwise integration of 1+\cos(2\mu)\leq 2\cos\mu.
最后还有一道类似的题目，可以打出你的方法和答案（其实不是很难）
（2009年湖北省预赛）设y = sin x + cos x + tan x + cot x + sec x + csc x，则| y |的最小值为__________________
