第二數(shù)學(xué)歸納法
顯然當(dāng)n=2時(shí)有
(x1+x2)/2≥(x1x2)^(1/2)
設(shè)當(dāng)n=1,2,...,k時(shí)成立,當(dāng)n=k+1時(shí)
則[x1+x2+...+xk+x(k+1)]+(k-1)(x1*x2*...x(k+1)^[1/(1+k)]
=(x1+x2+...+xk)+X(k+1)+(k-1)(x1*x2*...x(k+1)^[1/(1+k)]
≥k（x1*x2*...xk)^(1/k)+k{X(k+1)*[x1*x2*...x(k+1)]^[(k-1)/(k+1)]}(1/k)
≥k(2（x1*x2*...xk)^(1/2k)*{X(k+1)*[x1*x2*...x(k+1)]^[(k-1)/(k+1)]}(1/2k)
=2k(x1*x2*...x(k+1)^[1/(1+k)]
=> (x1+x2+...+xk+X(k+1)≥(k+1)(x1*x2*...x(k+1)^[1/(1+k)]
故當(dāng)n=k+1時(shí)也成立.原命題得證