Question

若xyz=1,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥3/2

Answer 1

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)