\(1=a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Rightarrow2P=2a^2+2b^2+2c^2=\frac{2}{a+b+c}+2ab+2bc+2ca\)
\(\Rightarrow3P=3a^2+3b^2+3c^2=\frac{2}{a+b+c}+a^2+b^2+c^2+2ab+2bc+2ca\)
\(=\frac{1}{a+b+c}+\frac{1}{a+b+c}+\left(a+b+c\right)^2\ge3\sqrt[3]{\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}}=3\)
\(\Rightarrow P\ge1\)
Đẳng thức xảy ra khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị.