Question

证明不等式 x+y+z=1,求证(x/y+1)+( y/z+1)+(z/x+1)≥3／4

Answer 1

记a=x/(y+1)+y/(z+1)+z/(x+1)；
因为上式是轮换对称式,故有：
a=z/(y+1)+x/(z+1)+y/(x+1)；两式相加得：
2a=（x+z）/(y+1)+(x+y)/(z+1)+(y+z)/(x+1)
=(1-y)/(1+y)+(1-z)/(1+z)+(1-x)/(1+x)
=2/(x+1)+2/(y+1)+2/(z+1)-3；
记1/(x+1)+1/(y+1)+1/(z+1)为m；并
应用均值不等式有：((x+1)+(y+1)+(z+1))/3≥3/(1/(x+1)+1/(y+1)+1/(z+1))〔算术平均≥调和平均〕上式即4/3≥3/m,
所以m≥9/4,往上回代,则有2a≥2*9/4-3=3/2,即a≥3/4,原式得证
P.S楼主括号打的有问题