数学归纳法证明不等式(1/n+1)+(1/n+2)+.+(1/3n+1)＞25/24

证明：n=1时,由 1/2+1/3+1/4 = 13/12 = 26/24 > 25/24知不等式成立.
现在设n = k的时候不等式成立,即 1/(k+1) + 1/(k+2) +...+1/(3k+1) > 25/24.①
则n = k+1时,
由 (3k+2)(3k+4) = (3k+3-1)(3k+3+1) = (3k+3)² - 1< (3k+3)²
知 {(3k+2)+(3k+4)}/{(3k+2)(3k+4)} > {(3k+2)+(3k+4)}/{ (3k+3)²}
即 1/(3k+2) + 1/(3k+4) > 2/(3k+3)
从而 1/(3k+2) +1/(3k+3) 1/(3k+4) > 3/(3k+3) = 1/(k+1) ②
因此有
　　1/(n+1)+1/(n+2)+...+1/(3n+1)
=　1/(k+2)+1/(k+3)+...+1/(3k+1)+1/(3k+2)+1/(3k+3)+1/(3k+4)
>　1/(k+2)+1/(k+3)+...+1/(3k+1) + 1/(k+1) .因为②
=　1/(k+1)＋1/(k+2)+1/(k+3)+...+1/(3k+1)
＞　25/24 .因为①
从而n=k+1时不等式成立.
因此由数学归纳法知原不等式对一切正整数n成立.