可用數(shù)學(xué)歸納法證明：
當(dāng)n=2時(shí),左邊=1+1/2+1/3+1/4=25/12,右邊=(7×2+11)12=25/12.原不等式成立.
假設(shè)n=k(k≥2)時(shí)命題成立.即1+1/2+1/3+…+1/2^k≥(7k+11)/12
那么,當(dāng)n=k+1時(shí)
1+1/2+1/3+…+1/2^(k+1)≥(7k+11)/12+1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)
而1/(2^k+1)+1/(2^k+2)+…+1/2^(k+1)
=1/(2^k+1)+1/(2^k+2)+…+1/[2^k+2^(k-1)]+1/[(2^k+2^(k-1)+1]+1/[(2^k+2^(k-1)+2]+…+1/2^(k+1)
≥2^(k-1)/[2^k+2^(k-1)]+2^(k-1)/2^(k+1)=7/12
于是,1+1/2+1/3+…+1/2^(k+1)≥(7k+11)/12+7/12=[7(k+1)+11]/12
所以,n=k+1時(shí)命題成立.
故對(duì)于一切n≥2,不等式1+1/2+1/3+...+1/2^n≥(7n+11)/12 都成立.