Áp dụng cosi:
\(\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\ge2ab+2bc+2ac\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow3\left(ab+bc+ac\right)\le\left(a+b+c\right)^2\)
\(\Leftrightarrow ab+bc+ac\le\dfrac{1}{3}\)
\(A=\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}+\dfrac{1}{2ab}+\dfrac{1}{2bc}+\dfrac{1}{2ca}+\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)
Áp dụng svac-xơ:
\(A\ge\dfrac{\left(1+1+1+1+1+1\right)^2}{a^2+b^2+c^2+3ab+3bc+3ac}+\dfrac{1}{2}.\dfrac{\left(1+1+1\right)^2}{ab+bc+ac}\)\(=\dfrac{36}{\left(a+b+c\right)^2+\left(ab+bc+ac\right)}+\dfrac{9}{2\left(ab+bc+ac\right)}\)\(\ge\dfrac{36}{1+\dfrac{1}{3}}+\dfrac{9}{\dfrac{2}{3}}\)
\(\Rightarrow A\ge\dfrac{81}{2}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Vậy \(minA=\dfrac{81}{2}\)
