[tex]\frac{1}{a^2+b^2+c^2}+\frac{1}{9abc}+\frac{8}{9abc}\geqslant \frac{1}{a^2+b^2+c^2}+\frac{1}{3(bc+ac+ab)^2}+\frac{8}{9\frac{(a+b+c)^3}{27}}\geqslant 2\sqrt{\frac{1}{a^2+b^2+c^2}.\frac{1}{3(bc+ac+ab)^2}}+24\geqslant 2\sqrt{\frac{1}{3\frac{(a^2+b^2+c^2+2ab+2bc+2ac)^2}{27}}}+24=30[/tex]
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