若a>0,b>0,且a+b=1,求证:(a+1/a)(b+1/b)≥25/4急 !

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若a>0,b>0,且a+b=1,求证:(a+1/a)(b+1/b)≥25/4急 !

若a>0,b>0,且a+b=1,求证:(a+1/a)(b+1/b)≥25/4急 !
若a>0,b>0,且a+b=1,求证:(a+1/a)(b+1/b)≥25/4
急 !

若a>0,b>0,且a+b=1,求证:(a+1/a)(b+1/b)≥25/4急 !
(a+1/a)(b+1/b)
=ab+b/a+a/b+1/(ab)
=(a^2b^2+b^2+a^2+1)/(ab)
=[a^2b^2+(a+b)^2-2ab+1]/(ab)
=[a^2b^2+1-2ab+1]/(ab)
=a^2b^2/ab-2ab/ab+2/ab
=ab+2/ab-2
1=a+b>=2√(ab)
所以√(ab)