求证x/(y+z-x)+y/(x+z-y)+z/(x+y-z)≥3

x/(y+z-x)+y/(x+z-y)+z/(x+y-z)≥3
令y+z-x=a,x+z-y=b,x+y-z=c
z=a+b/2 x=b+c/2 y=a+c/2
原式化为：
证明b+c/2a + a+c/2b+a+b/2c≥3
即证明b/a+c/a+a/b+c/b+a/c+b/c≥6
b/a+a/b+c/a+a/c+b/c+c/b≥6
好像给的条件不够啊，如果满足左边各项均大于0，下面接着用不等式定理好了！～
b/a+a/b+c/a+a/c+b/c+c/b≥2+2+2＝6

条件不够，举个反例子：x=0 y=0 z=1
代进去为－1 不满足

用换元,令y+z-x=a,x+z-y=b,x+y-z=c,式子左边都换成a,b,c的式子,题目迎刃而解.

支持小南VS仙子的反例！