已知正数a,b,c满足1/a+1/b+1/c=1,证明:a/(1+a)+b/(1+b)+c/(1+c)大于等于9/4、

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令1/a=x 1/b=y 1/c=z
则x+y+z=1
(1+x)+(1+y)+(1+z)=4 (1)
a/(1+a)+b/(1+b)+c/(1+c)=1/(1/a+1)+1/(1/b+1)+1/(1/c+1) 变为
1/(x+1)+1/(y+1)+1/(z+1)>=9/4
[1/(x+1)+1/(y+1)+1/(z+1)]((1+x)+(1+y)+(1+z))
=1+(1+y)/(1+x)+(1+z)/(1+x)+1+(1+x)/(1+y)+(1+z)/(1+y)+1+(1+x)/(1+z)+(1+y)/(1+z)
=3+[(1+y)/(1+x)+(1+x)/(1+y)]+[(1+z)/(1+y)+(1+y)/(1+z)]+[(1+z)/(1+x)+(1+x)/(1+z)]
>=3+6
=9
(1)=4
所以
1/(x+1)+1/(y+1)+1/(z+1)>=9/4
所以 a/(1+a)+b/(1+b)+c/(1+c)>=9/4

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