求证a／(b+c)+b/(a+c)+c/(a+b)≥3/2

令b+c=x,a+c=y,a+b=z
a=(y+z-x)/2,b=(x+z-y)/2,c=(x+y-z)/2
所以a/(b+c)+b/(a+c)+c/(a+b)
=(y+z-x)/2x+(x+z-y)/2y+(x+y-z)/2z
=[(y+z)/x +(x+z)/y +(x+y)/z]/2 -3/2
=[y/x +x/y +z/x +x/z +y/z +z/y]/2 -3/2
=[(y/x +x/y) +(z/x +x/z) +(y/z +z/y)]/2 -3/2
>=[2+2+2]/2 -3/2 {由于y/x +x/y>=2*根号(y/x * x/y)=2}
=3/2

左边=(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)-3
=0.5*(a+b+b+c+c+a)*[1/(a+b)+1/(b+c)+1/(c+a)]-3
≥0.5*{3*[(a+b)(b+c)(c+a)]^1/3}*{3*[1/(a+b)*1/(b+c)*1/(c+a)]^1/3}-3
=0.5*3*3-3=3/2

参考资料：http://zhidao.baidu.com/question/3152042.html?si=2

提示：用均值不等式解