Cho a,b,c>0 thỏa mãn a+b+c=1.CMR $\frac{1}{a^{2}+b^{2}+c^{2}}+\frac{1}{abc}\geq 30$
VT= $\frac{1}{1-2(ab+bc+ac)}+\frac{1}{abc}\geq \frac{1}{1-6\sqrt[3]{(abc)^2}}+\frac{1}{abc}=\frac{1}{1-6t^2}+\frac{1}{t^3}$
$=\frac{1}{1-6t^2}+\frac{1}{9t^3}+....+\frac{1}{9t^3}\geq \frac{100}{1-6t^2+81t^3}$
Có $\frac{100}{1-6t^2+81t^3}\geq 30 <=> 243t^3-18t^2-7\leq 0 <=> (3t-1)(81t^2+21t+7)\leq 0$ (luôn đúng)
$t=\sqrt[3]{abc}$ ($0< t\leq \frac{1}{3}$)