Sa se arate ca daca a,b,c sunt numere pozitive , atunci :
\( \sum_{cyc}\frac{a^3}{a^2+ab+b^2}\ge \frac{a+b+c}{3} \) .
Inegalitati elementare, Bogdan Enescu
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Mateescu Constantin
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Avem egalitatile: \( \left\|\ \begin{array}{ccc}
\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}} & = & a-b \\\\\\\\
\frac{b^{3}-c^{3}}{b^{2}+bc+c^{2}} & = & b-c \\\\\\\\
\frac{c^{3}-a^{3}}{c^{2}+ca+a^{2}} & = & c-a\ \end{array}\right|\ \bigoplus\ \Longrightarrow\ \sum_{cyc}\frac{a^{3}}{a^{2}+ab+b^{2}}=\sum_{cyc}\frac{b^{3}}{a^{2}+ab+b^{2}}=\frac{1}{2}\sum_{cyc}\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}. \)
Acum, \( \ \frac{1}{2}\cdot\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}\geq\frac{a+b}{6} \) si analoagele .
\( \Longrightarrow\ LHS\geq\frac{a+b+b+c+c+a}{6}=\frac{a+b+c}{3}. \)
\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}} & = & a-b \\\\\\\\
\frac{b^{3}-c^{3}}{b^{2}+bc+c^{2}} & = & b-c \\\\\\\\
\frac{c^{3}-a^{3}}{c^{2}+ca+a^{2}} & = & c-a\ \end{array}\right|\ \bigoplus\ \Longrightarrow\ \sum_{cyc}\frac{a^{3}}{a^{2}+ab+b^{2}}=\sum_{cyc}\frac{b^{3}}{a^{2}+ab+b^{2}}=\frac{1}{2}\sum_{cyc}\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}. \)
Acum, \( \ \frac{1}{2}\cdot\frac{a^{3}+b^{3}}{a^{2}+ab+b^{2}}\geq\frac{a+b}{6} \) si analoagele .
\( \Longrightarrow\ LHS\geq\frac{a+b+b+c+c+a}{6}=\frac{a+b+c}{3}. \)