\(P=\frac{2a}{2\sqrt{\left(b+1\right)\left(b^2-b+1\right)}+2}+\frac{2b}{2\sqrt{\left(c+1\right)\left(c^2-c+1\right)}+2}\)\(+\frac{2c}{2\sqrt{\left(a+1\right)\left(a^2-a+1\right)}+2}\)
\(P\ge\frac{2a}{b^2+4}+\frac{2b}{c^2+4}+\frac{2c}{a^2+4}\)
\(2P\ge\frac{4a}{b^2+4}+\frac{4b}{c^2+4}+\frac{4c}{a^2+4}=a-\frac{ab^2}{b^2+4}+b-\frac{bc^2}{c^2+4}+a-\frac{ca^2}{a^2+4}\)
\(2p\ge a+b+c-\left(\frac{ab^2}{4b}+\frac{bc^2}{4c}+\frac{ca^2}{4a}\right)\)
\(2P\ge6-\frac{1}{4}\left(ab+bc+ca\right)\ge6-\frac{1}{12}\left(a+b+c\right)^2=3\)
\(\Rightarrow P\ge\frac{3}{2}\)
Dấu " = " xảy ra khi \(a=b=c=2\)
