Исбот кунед, ки

\(1+\frac{1}{2}+...+\frac{1}{n}\geq n(\sqrt[n]{n+1}-1)\)

\(S_n = 1+\frac{1}{2}+...+\frac{1}{n}\)

\(S_n\geq n(\sqrt[n]{n+1}-1)\)

\(S_n\geq n\cdot\sqrt[n]{n+1}-n\)

\(S_n+n\geq n\cdot\sqrt[n]{n+1}\)

Акнун мо бояд исбот кунем, ки \(S_n+n\geq n\cdot\sqrt[n]{n+1}.\)

\(S_n+n = n+(1+\frac{1}{2}+...+\frac{1}{n})=\)

\(=(1+1)+(1+\frac{1}{2})+...+(1+\frac{1}{n})=\)

\(=\frac{2}{1}+\frac{3}{2}+...+\frac{n+1}{n}=\)

\(=n\cdot(\frac{\frac{2}{1}+\frac{3}{2}+...+\frac{n+1}{n}}{n})\)

\(\frac{a_1+a_2+...+a_n}{n}\geq\sqrt[n]{a_1\cdot a_2\cdot..\cdot a_n}\)

\(\frac{\frac{2}{1}+\frac{3}{2}+...+\frac{n+1}{n}}{n}\geq\sqrt[n]{\frac{2}{1}\cdot\frac{3}{2}\cdot...\cdot\frac{n+1}{n}}=\)

\(=\sqrt[n]{n+1}\)

Яъне,

\(n\cdot(\frac{\frac{2}{1}+\frac{3}{2}+...+\frac{n+1}{n}}{n})\geq n\cdot\sqrt[n]{n+1}\)

\(S_n+n\geq n\cdot\sqrt[n]{n+1}\)

\(S_n\geq n\cdot\sqrt[n]{n+1}-n\)

\(S_n\geq n\cdot(\sqrt[n]{n+1}-1)\)

Исбот шуд.