Inégalité (24/11/2006)

Soit 0 < a <= x1 <= x2 <= ... <= xn <= b.
Montrer que
(x1+x2+...+xn)(1/x1+1/x2+...+1/xn)<=((a+b)²/4ab)n².

on pose m=1/rac(ab)
ona 4sum(x_i)sum(1/x_i)=4(sum(mx_i))(sum(1/mx_i))<=(msum(x_i)+(1/m)sum(1/x_i))^2<=n^2(mb+1/ma)^2=n^2(a+b)^2/ab