\(P=\dfrac{a^2+b^2+c^2+2\left(ab+bc+ca\right)+1}{a+b+c-abc}=\dfrac{\left(a+b+c\right)^2+1}{a+b+c-abc}\ge\dfrac{\left(a+b+c\right)^2+1}{a+b+c}\)
\(\Rightarrow P\ge a+b+c+\dfrac{1}{a+b+c}\) (1)
\(P=\dfrac{a^2+b^2+c^2+3\left(ab+bc+ca\right)}{\left(a+b+c\right)\left(ab+bc+ca\right)-abc}=\dfrac{\left(a+b\right)\left(b+c\right)+\left(b+c\right)\left(c+a\right)+\left(a+b\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(P=\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{a+b+c}\left(\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+c+b}{a+c}\right)\)
\(P=\dfrac{1}{a+b+c}\left(3+\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\ge\dfrac{1}{a+b+c}\left(3+\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\right)\)
\(P\ge\dfrac{1}{a+b+c}\left(3+\dfrac{\left(a+b+c\right)^2}{2}\right)=\dfrac{3}{a+b+c}+\dfrac{a+b+c}{2}\)
\(\Rightarrow3P\ge\dfrac{3}{2}\left(a+b+c\right)+\dfrac{9}{a+b+c}\) (2)
Cộng vế (1) và (2):
\(\Rightarrow4P\ge\dfrac{5}{2}\left(a+b+c\right)+\dfrac{10}{a+b+c}\ge2\sqrt{\dfrac{50\left(a+b+c\right)}{2\left(a+b+c\right)}}=10\)
\(\Rightarrow P\ge\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;1;0\right)\) và các hoán vị
