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Daca a,b,c sunt pozitive si \( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3 \) atunci \( \frac{a^2+1}{\sqrt{a^2-a+1}}+\frac{b^2+1}{\sqrt{b^2-b+1}}+\frac{c^2+1}{\sqrt{c^2-c+1}}\ge 6 \)

,,Recreatii Matematice''

Cum \( 3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\Longrightarrow abc\ge1
\) si \( \frac{a^{2}+1}{\sqrt{a^{2}-a+1}}=\frac{a^{2}-a+1+a}{\sqrt{a^{2}-a+1}}=\sqrt{a^{2}-a+1}+\frac{a}{\sqrt{a^{2}-a+1}} \) rezulta imediat cerinta:

\( \sum\frac{a^{2}+1}{\sqrt{a^{2}-a+1}}=\sum\sqrt{a^{2}-a+1}+\sum\frac{a}{\sqrt{a^{2}-a+1}}\ge6\sqrt[6]{abc}\ge6 \).

Aceasta este inegalitatea propusa I.V. Maftei din Shortlist ONM 2006, clasa a IX-a.