Dễ dàng chứng minh được BĐT phụ: \(\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)
\(P=\dfrac{a^2}{2ab+3ac}+\dfrac{b^2}{2bc+3ab}+\dfrac{c^2}{2ac+3bc}\ge\dfrac{\left(a+b+c\right)^2}{5\left(ab+ac+bc\right)}\)
\(P\ge\dfrac{\left(a+b+c\right)^2}{\dfrac{5}{3}.3\left(ab+ac+bc\right)}\ge\dfrac{\left(a+b+c\right)^2}{\dfrac{5}{3}\left(a+b+c\right)^2}=\dfrac{3}{5}\)
\(\Rightarrow P_{min}=\dfrac{3}{5}\) khi \(a=b=c\)
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