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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
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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
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【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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