\(P=\sum\frac{a^3}{b^2+ab+bc+ca}=\sum\frac{a^3}{\left(a+b\right)\left(b+c\right)}\)
\(\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3a}{4}\)
Tương tự và cộng lại:
\(P+\frac{1}{2}\left(a+b+c\right)\ge\frac{3}{4}\left(a+b+c\right)\Rightarrow P\ge\frac{1}{4}\left(a+b+c\right)\)
\(\Rightarrow P\ge\frac{1}{4}\sqrt{3\left(ab+bc+ca\right)}=\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)