已知x,y,z都是正数,且xyz=1,求证:x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥3/2
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已知x,y,z都是正数,且xyz=1,求证:x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥3/2
柯西
【x^2/(y+z)+y^2/(x+z)+z^2/(x+y)】*
(y+z+x+z+x+y)≥(x+y+z)^2
即
x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥(x+y+z)/2=(3/2)(x+y+z)/3≥(3/2)(xyz)^1/3=3/2
【x^2/(y+z)+y^2/(x+z)+z^2/(x+y)】*
(y+z+x+z+x+y)≥(x+y+z)^2
即
x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥(x+y+z)/2=(3/2)(x+y+z)/3≥(3/2)(xyz)^1/3=3/2
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