Cho 3 số dương a,b,c thỏa mãn a +b+c>=1tìm GTNN của bt P=a^3/3a^2+b +b^3/3b^2 +c +c^3/3c^2+a
\[\begin{array}{l}
P = \frac{{{a^3}}}{{3{a^2} + b}} + \frac{{{b^3}}}{{3{b^2} + c}} + \frac{{{c^3}}}{{3{c^2} + a}}\\
\frac{{{a^3}}}{{3{a^2} + b}} = \frac{a}{3} - \frac{{ab}}{{3\left( {3{a^2} + b} \right)}} \ge \frac{a}{3} - \frac{{ab}}{{3*2\sqrt {3{a^2}b} }} = \frac{a}{3} - \frac{{\sqrt {3b} }}{{18}}(3{a^2} + b \ge 2\sqrt {3{a^2}b} \to \frac{1}{{3{a^2} + b}} \le \frac{1}{{2\sqrt {3{a^2}b} }} \to - \frac{1}{{3{a^2} + b}} \ge - \frac{1}{{2\sqrt {3{a^2}b} }})\\
tuong.tu\\
\frac{{{b^3}}}{{3{b^2} + c}} \ge \frac{b}{3} - \frac{{\sqrt {3c} }}{{18}}\\
\frac{{{c^3}}}{{3{c^2} + a}} \ge \frac{c}{3} - \frac{{\sqrt {3a} }}{{18}}\\
\to P \ge \frac{1}{{18}}\left( {6a + 6b + 6c - \sqrt {3a} - \sqrt {3b} - \sqrt {3c} } \right)\\
\frac{{3a + 1}}{2} \ge \sqrt {3a} \leftrightarrow \frac{{3a}}{2} - \sqrt {3a} \ge \frac{{ - 1}}{2}\\
\frac{{3b + 1}}{2} \ge \sqrt {3b} \leftrightarrow ..\\
\frac{{3c + 1}}{2} \ge \sqrt {3c} \leftrightarrow ...\\
\to P \ge \frac{1}{{18}}\left( {\frac{9}{2}\left( {a + b + c} \right) + \frac{3}{2}\left( {a + b + c} \right) - \sqrt {3a} - \sqrt {3b} - \sqrt {3c} } \right) \ge \frac{1}{{18}}\left( {\frac{9}{2}\left( {a + b + c} \right) - \frac{3}{2}} \right) \ge \frac{1}{{18}}\left( {\frac{9}{2} - \frac{3}{2}} \right) = \frac{1}{6}\\
dau = \leftrightarrow a = b = c = \frac{1}{3}
\end{array}\]
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