解:a、b均为正数
1、∵(a-b)²≥0
=>a²+b²≥2ab
=>√[(a²+b²)/2]=√[(a²+b²+a²+b²)/4]≥√[(a²+b²+2ab)/4]=√(a+b)²=a+b
即:√[(a²+b²)/2]≥a+b
2.∵(√a-√b)²≥0
=>a+b≥2√ab
=>(a+b)/2≥√ab
3.2/(1/a+1/b)=2ab/(a+b)
又∵a+b≥2√ab
=>2/(1/a+1/b)=2ab/(a+b)≤2ab/2√ab=√ab
即:√ab≥2/(1/a+1/b)
综合得出:√[(a²+b²)/2]≥(a+b)/2≥√ab≥2/(1/a+1/b) (a,b均为正数)
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