若xyz=1,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥3/2
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若xyz=1,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥3/2
若xyz=1,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥3/2
若xyz=1,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥3/2
x/(y+z)+y/(x+z)+z/(x+y)>=3/2
设S=x+y+z
x/(y+z)+y/(x+z)+z/(x+y)
=S/(y+z)+S/(x+z)+S/(x+y)-3
>=9/[(y+z)/S+(x+z)/S+(y+x)/S]-3
=9/2-3
=3/2
以上不等号是用算术平均>=调和平均,即:a+b+c/3>=3/(1/a+1/b+1/c)
变一下就是a+b+c>=9/(1/a+1/b+1/c)
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