a>0 b>0,求证(a+b)(a^-1+b^-1)>=4
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![a>0 b>0,求证(a+b)(a^-1+b^-1)>=4](/uploads/image/z/13620181-13-1.jpg?t=a%3E0+b%3E0%2C%E6%B1%82%E8%AF%81%EF%BC%88a%2Bb%29%28a%5E-1%2Bb%5E-1%29%3E%3D4)
a>0 b>0,求证(a+b)(a^-1+b^-1)>=4
a>0 b>0,求证(a+b)(a^-1+b^-1)>=4
a>0 b>0,求证(a+b)(a^-1+b^-1)>=4
(a+b)(1/a+1/b)
=1+a/b+b/a+1
=2+(a/b+b/a)
a/b>0,b/a>0
所以a/b+b/a≥2√(a/b*b/a)=2
所以(a+b)(1/a+1/b)≥2+2
即(a+b)(1/a+1/b)≥4