\(\|x\| + \|y\| \ge \|x + y\|\)
using convexity of norm to prove it.
we need to use \(\|\alpha\|=\alpha\) among the process, where \alpha is a constant.
\(\|\alpha x\| + \|(1-\alpha) y\|\ge \|\alpha x+(1-\alpha )y\|\), let\(\alpha=1/2\), we get
\(\|\frac{1}{2} x\| + \|\frac{1}{2}y\| \ge \|\frac{1}{2}(x + y)\|\) \(\Rightarrow \frac{1}{2} (\|x\| + \|y\|)\ge \frac{1}{2}\|(x + y)\|\)
thus \(\|x\| + \|y\| \ge \|x + y\|\)
标签：among,Triangle,frac,ge,inequality,alpha From： https://www.cnblogs.com/xytang-mini-juan/p/17151760.html