已知x>0,y>0,z>0,求x/(y+z)+y/(x+z)+z/(x+y)>=3/2

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)

2: 若x>0,y>0,z>0 且√x＋√y≤2a√(x+y)
a>= (√x＋√y)/[2*√(x+y)]
而:(√x＋√y)/2<=√[(x+y)/2](算术平均<=几何平均)
(√x＋√y)<=√(x+y)*√2
(√x＋√y)/[2*√(x+y)]<=√2/2
√x＋√y≤2a√(x+y)恒成立
a>=√2/2,a最小为√2/2

已知x>0,y>0,z>0,求证：x/(y+z)+y/(x+z)+z/(x+y)≥3/2
证明：
忘了！