\(P=\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ca}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ca+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5\left(ab+bc+ca\right)}\) (BĐT Bunhiacopxki dạng phân thức)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ca\right)}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}\) (vì \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\) )
\(=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Vậy \(MinP=\dfrac{1}{2}\)
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