就像标题所说，我们来一道高联题目练练手

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（请注意，这个结论不可以直接使用，因为这个也是高联证明题之一！将在下一篇文章具体证明！）

$\frac{x}{\sin x} = \frac{x}{2\sin \frac{x}{2} \cos \frac{x}{2}} = \frac{x}{4\sin \frac{x}{4}\cos\frac{x}{4} \cos \frac{x}{2}} = \frac{\frac{x}{4}}{\tan \frac{x}{4} \cos^2\frac{x}{4}\cos\frac{x}{2}} \le \frac{1}{\cos^2\frac{x}{4}\cos\frac{x}{2}}$

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 $1+\cos(2\mu)\leq 2\cos\mu$.

最后再送大家一道题目吧，大家可以在评论区打出你的方法哦~

（2009年湖北省预赛）设$y = \sin x + \cos x + \tan x + \cot x + \sec x + \csc x$，则$\left | y \right |$的最小值为__________________

答案：$2\sqrt{2} – 1$