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(a²+b²)/√ab≥a+b（其中a>0 b>0）怎么证明啊
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(a²+b²)/√ab≥a+b（其中a>0 b>0）怎么证明啊
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答案和解析
令：√a=x,√b=y,则：
(a²+b²)/√ab-(a+b)
=(x^4+y^4)/(xy)-(x^2+y^2)
=(x^4+y^4-x^3y-xy^3)/(xy)
=[x^3(x-y)-y^3(x-y)]/(xy)
=(x-y)^2(x^2+xy+y^2)/(xy)
∵(x-y)^2≥0,(x^2+xy+y^2)>0,(xy)>0
∴(a²+b²)/√ab-(a+b)≥0
(a²+b²)/√ab≥(a+b)
(a²+b²)/√ab-(a+b)
=(x^4+y^4)/(xy)-(x^2+y^2)
=(x^4+y^4-x^3y-xy^3)/(xy)
=[x^3(x-y)-y^3(x-y)]/(xy)
=(x-y)^2(x^2+xy+y^2)/(xy)
∵(x-y)^2≥0,(x^2+xy+y^2)>0,(xy)>0
∴(a²+b²)/√ab-(a+b)≥0
(a²+b²)/√ab≥(a+b)
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