Sử dụng BĐT quen thuộc: \(\frac{8}{9}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Rightarrow P\ge3\left(a^2+b^2+c^2\right)+\frac{24\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b+c\right)^3}\)
\(P\ge3\left(a^2+b^2+c^2\right)+\frac{192}{\left(a+b+c\right)^2}\)
\(P\ge\left(a+b+c\right)^2+\frac{192}{\left(a+b+c\right)^2}\)
\(P\ge\frac{\left(a+b+c\right)^2}{3}+\frac{192}{\left(a+b+c\right)^2}+\frac{2\left(a+b+c\right)^2}{3}\)
\(P\ge2\sqrt{\frac{192\left(a+b+c\right)^2}{3\left(a+b+c\right)^2}}+2\left(ab+bc+ca\right)=32\)
Dấu "=" xảy ra khi \(a=b=c=\frac{2\sqrt{6}}{3}\)
