问题: 已知abcd都是正实数.设s=a/(a+b+c)+b/(b+c+d)+c/(c+d+a)+d/(d+a+b) 则怎样证明1<S<2

s=a/(a+b+c)+b/(b+c+d)+c/(c+d+a)+d/(d+a+b)>a/(a+b+c+d)+b/(a+b+c+d)+c/(a+b+c+d)+d/(a+b+c+d)=(a+b+c+d)/(a+b+c+d)=1 s=a/(a+b+c)+b/(b+c+d)+c/(c+d+a)+d/(d+a+b)=4-(b+c)/(a+b+c)-(c+d)/(b+c+d)-(d+a)&#