給你兩種證法:
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(順序和)
≥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(亂序和)
=x^n+x^n+x^n+...+x^n
=(2n+1)x^n
等號(hào)成立當(dāng)且僅當(dāng)x=1
2.用基本不等式,算術(shù)平均≥幾何平均
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
等號(hào)成立當(dāng)且僅當(dāng)x=1