Question

求证(x1+x2+...xn)^2/2(x1^2+x2^2+.xn^2)≤x1/(x2+x3)+x2/(x3+x4)+.？

Answer 1

证明： 由排序不等式, x1^2+x2^2+...+xn^2>=x1x2+x2x3+...xn-1xn+xnx1 x1^2+x2^2+...+xn^2>=x1x3+x2x4+...xn-1x1+xnx2 两式相加得 2(x1^2+x2^2+...+xn^2)>=x1(x2+x3)+x2(x3+x4)+...+xn-1(xn+x1)+xn(x1+x2) 又因为由柯西不等式 [x1/(x2+x3)+x2/(x3+x4)+...+xn/(x1+x2)]*[x1(x2+x3)+x2(x3+x4)+...+xn(x1+x2)] >=(x1+x2+...+xn)^2 所以 [x1/(x2+x3)+x2/(x3+x4)+...+xn/(x1+x2)]*2(x1^2+x2^2+...+xn^2) >=(x1+x2+...+xn)^2 即 (x1+x2+...xn)^2/2(x1^2+x2^2+...+xn^2),2,求证(x1+x2+...xn)^2/2(x1^2+x2^2+.xn^2)≤x1/(x2+x3)+x2/(x3+x4)+.xn/(x1+x2)