Chỉ nghĩ được thế này, hơi dài dòng văn tự:
\(P\ge\frac{\left(a+b+c\right)^2}{a^2b+b^2c+c^2a+ab+bc+ca}\)
Mặt khác:
\(3\left(a+b+c\right)=\left(a^2+b^2+c^2\right)\left(a+b+c\right)\)
\(=a^3+ab^2+b^3+bc^2+c^3+ca^2+a^2b+b^2c+c^2a\)
\(\ge2a^2b+2b^2c+2c^2a+a^2b+b^2c+c^2a=3\left(a^2b+b^2c+c^2a\right)\)
\(\Rightarrow a+b+c\ge a^2b+b^2c+c^2a\)
\(\Rightarrow P\ge\frac{\left(a+b+c\right)^2}{a+b+c+ab+bc+ca}\)
Dễ dàng chứng minh \(\frac{\left(a+b+c\right)^2}{a+b+c+ab+bc+ca}\ge\frac{3}{2}\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+ab+bc+ca\ge3\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2-6\left(a+b+c\right)+3\left(a^2+b^2+c^2\right)\ge0\)
\(\Leftrightarrow\left(a+b+c-3\right)^2\ge0\)
