Question

排序不等式证明设x>0,求证1+x+x^2+x^3+.+x^(2n)>=(2n+1)*x^n实在用排序证不出来用其他的也

Answer 1

给你两种证法:

1.用排序不等式:

1+x+x^2+...+x^2n

=x^0*x^0+x^(1/2)*x^(1/2)+x^1*x^1+...x^(n-2/2)*x(n-2/2)+x^(n-1/2)*x^(n-1/2)+x^n*x^n(顺序和)

≥x^0*x^n+x^(1/2)*(x^(n-1/2)+x^1*x^(n-2/2)+...+x^(n-1/2)*x^(1/2)+x^n*x^0(乱序和)

=x^n+x^n+x^n+...+x^n

=(2n+1)x^n

等号成立当且仅当x=1

2.用基本不等式,算术平均≥几何平均

1+x+x^2+...+x^n

≥(2n+1)(1*x*x^2*..*x^2n)(1/(2n+1))

=(2n+1)(x^(2n+1)*n)^(1/(2n+1))

=(2n+1)x^n

等号成立当且仅当x=1