请问第四小题是怎么做出来≥4的 谢谢

Question

如题所述

Answer 1

原式=c/a+b/d+d/b+a/c
=(c/a+a/c)(b/d+d/b)
=(a²+c²)/ac+(b²+d²)/bd
=(a²+c²-2ac+2ac)/ac+(b²+d²-2bd+2bd)/bd
=[(a-c)²+2ac]/ac+[(b-d)²+2bd]/bd
=(a-c)²/ac+2+(b-d)²bd+2
=(a-c)²/ac+(b-d)²/bd+4
∵a-c)²/ac+(b-d)²/bd≥0
∴(a-c)²/ac+(b-d)²/bd+4≥4追问

(⊙o⊙)哦 原来是这样 谢谢！