\(A=\Sigma\dfrac{1}{a^5\left(b+2c\right)^2}=\Sigma\dfrac{\left(abc\right)^3}{a^5\left(b+2c\right)^2}=\Sigma\dfrac{b^3c^3}{a^2\left(b+2c\right)^2}=\Sigma\dfrac{b^3c^3}{\left(ab+2ac\right)^2}\)
\(có:\dfrac{b^3c^3}{\left(ab+2ac\right)^2}+\dfrac{ab+2ac}{27}+\dfrac{ab+2ac}{27}\ge3\sqrt[3]{\dfrac{b^3c^3}{27^2}}=\dfrac{bc}{3}\)
\(tương\) \(tự\Rightarrow A\ge\left(\dfrac{bc+ac+ab}{3}\right)-\left(\dfrac{2\left(ab+2ac\right)+2\left(bc+2ab\right)+2\left(ca+2bc\right)}{27}\right)\)
\(=\dfrac{9\left(ab+bc+ac\right)-2\left(ab+2ac+bc+2ab+ca+2bc\right)}{27}=\dfrac{3\left(ab+bc+ca\right)}{27}\ge\dfrac{3}{27}.3\sqrt[3]{\left(abc\right)^2}=\dfrac{1}{3}\)
\(dấu"="\Leftrightarrow a=b=c=1\)
