$ 6P=\sum \ \frac{6}{a^2 + b^2} - \frac{a^3 + b^3 + c^3}{2abc} $
$ = 3 + \sum \frac{c^2}{a^2+b^2} - \frac{a^3 + b^3 + c^3}{2abc} $
$ \leq 3 + \sum \frac{c^2}{2ab} - \frac{a^3 + b^3 + c^3}{2abc} $
$ = 3 + \frac{a^3 + b^3 + c^3}{2abc} - \frac{a^3 + b^3 + c^3}{2abc}
=3 $
$ => P \leq \frac{1}{2} Dau = xay ra \leftrightarrow a=b=c=\sqrt{2} $
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#739766 $P= \sum\frac{1}{a^{2}+b^{2...
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on 01-06-2023 - 11:54
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Bất đẳng thức và cực trị
