Cho a b c >0 và a+b=c=1. Tìm min F=$\frac{9}{1-2(ab+ac+bc)}+\frac{2}{abc}$
ta cos $\frac{9}{1-2(ab+bc+ac)}=\frac{9}{(a+b+c)^{2}-2(ab+bc+ac)}=\frac{9}{a^{2}+b^{2}+c^{2}}$
conf $\frac{2}{abc}=\frac{2(a+b+c)}{abc}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\geq \frac{2.9}{ab+bc+ac}$
vt $\geq \frac{9}{a^{2}+b^{2}+c^{2}}+\frac{9}{ab+bc+ac}+\frac{9}{ab+bc+ac}\geq \frac{(3+3+3)^{2}}{(a+b+c)^{2}}=81$
dau =sayr ra khi a=b=c=1/3
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