Исбот кунед, ки \(\frac{1}{a-2}+\frac{1}{a-1}+\frac{1}{a+1}+\frac{1}{a+2}>\frac{4}{a}\), агар a>2 бошад.

\(a>2\), яъне \(\frac{1}{a-2}>0\), \(\frac{1}{a-1}>0\),

\(\frac{1}{a+1}>0\), \(\frac{1}{a+2}>0\)

\(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\geq\frac{n^2}{a_1+a_2+...+a_n}\)

\(\frac{1}{a-2}+\frac{1}{a-1}+\frac{1}{a+1}+\frac{1}{a+2}\geq\frac{16}{a-2+a-1+a+1+a+2}=\)

\(=\frac{16}{4\cdot a}=\frac{4}{a}\)

Нишон додан мондааст, ки \(\frac{1}{a-2}+\frac{1}{a-1}+\frac{1}{a+1}+\frac{1}{a+2}\neq\frac{4}{a}\)

Инро аз рӯи нобаробарии зерин нишон додан мумкин аст:

\(\frac{1}{a-2}\neq\frac{1}{a-1}\neq\frac{1}{a+1}\neq\frac{1}{a+2}\)

Исбот шуд.