Neka je [inlmath]n[/inlmath] prirodni broj. Dokazi nejednakost:
[dispmath]\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n+1}>1[/dispmath] Imam ideju da se radi preko indukcije, ali je ne znam bas realizovati. Pomoc je dobrodosla!

Baza [inlmath]T\left( n=1 \right)[/inlmath]: [dispmath]\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1[/dispmath]Hipoteza [inlmath]T\left( n=k \right)[/inlmath]:
[dispmath]\frac {1}{k+1}+\frac {1}{k+2}+...+\frac {1}{3k+1}>1[/dispmath]Induktivni dokaz [inlmath]T\left( n=k+1 \right)[/inlmath] (potrebno je dokazati):
[dispmath]\frac {1}{k+2}+\frac {1}{k+3}+...+\frac {1}{3\left( k+1 \right)+1}>1[/dispmath]Krenimo od leve strane nejednakosti:
[dispmath]\frac {1}{k+2}+\frac {1}{k+3}+...+\frac {1}{3\left( k+1 \right)+1}=[/dispmath] [dispmath]=\underbrace{\frac {1}{k+1}+\frac {1}{k+2}+\frac {1}{k+3}+...+\frac {1}{3k+1}}_{hipoteza: >1}+\frac {1}{3k+2}+\frac {1}{3k+3}+\frac {1}{3k+4} - \frac {1}{k+1} {\color{red}>}[/dispmath] [dispmath]>1+\frac {1}{3k+2}+\frac {1}{3\left(k+1 \right) }+\frac {1}{3k+4} - \frac {1}{k+1} =[/dispmath] [dispmath]=1+\frac {1}{3k+2}-\frac {2}{3\left(k+1 \right) }+\frac {1}{3k+4} =[/dispmath] [dispmath]=1+\frac {3\left(k+1 \right)\left(3k+4 \right) - 2\left(3k+2 \right)\left(3k+4 \right) + 3\left(k+1 \right)\left(3k+2 \right)}{3\left(k+1 \right)\left(3k+2 \right)\left(3k+4 \right)} =[/dispmath] [dispmath]=1+\frac {9k^2+21k+12-18k^2-36k-16+9k^2+15k+6}{3\left(k+1 \right)\left(3k+2 \right)\left(3k+4 \right)} =[/dispmath] [dispmath]=1+\frac {2}{3\left(k+1 \right)\left(3k+2 \right)\left(3k+4 \right)} {\color{red}>} 1[/dispmath]Poslednja nejednakost je očigledna jer je [inlmath]k>0[/inlmath].