设0

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设0设0设0a/x=(1-x+x)a/x=(1-x)a/x+x*a/x=(1-x)a/x+ab/(1-x)=(1-x+x)b/(1-x)=(1-x)b(1-x)+x*b/(1-x)00所以(1-x)a

设0
设0

设0
a/x=(1-x+x)a/x =(1-x)a/x+x*a/x =(1-x)a/x+a b/(1-x) =(1-x+x)b/(1-x) =(1-x)b(1-x)+x*b/(1-x) 00 所以(1-x)a/x>0,x*b/(1-x)>0 所以(1-x)a/x+x*b/(1-x)>=2√[(1-x)a/x*x*b/(1-x)]=2ab 所以原式>=2ab+a+b 所以最小值=(a+b)