Question

不等式证明速进!如何证明bc/(a^2(b+c))+ac/(b^2(a+c))+ba/(c^2(b+a))>=1/2(1

Answer 1

a,b,c都是正数吧.

证明：

令a=1/x,b=1/y,c=1/z

那么原不等式即为：

x^2/(y+z)+y^2/(x+z)+z^2/(x+y)>=(1/2)(x+y+z)

根据柯西不等式：

[(y+z)+(x+z)+(x+y)][x^2/(y+z)+y^2/(x+z)+z^2/(x+y)]>=(x+y+z)^2

∴2(x+y+z)[x^2/(y+z)+y^2/(x+z)+z^2/(x+y)]>=(x+y+z)^2

即：x^2/(y+z)+y^2/(x+z)+z^2/(x+y)>=(1/2)(x+y+z)

等号当且仅当x=y=z时成立.

故原不等式得证,当且仅当a=b=c时取等号.