Ta có: \(P=\Sigma\dfrac{a^2\left(b+1\right)}{a\left(b+1\right)+b}=\Sigma\dfrac{a^2\left(b+1\right)+ab-ab}{a\left(b+1\right)+b}=\Sigma\left(a-\dfrac{ab}{a\left(b+1\right)+b}\right)\)
\(\Rightarrow P=\left(a+b+c\right)-\Sigma\dfrac{ab}{a\left(b+1\right)+b}=3-\Sigma\dfrac{ab}{a\left(b+1\right)+b}\)
Áp dụng BĐT Cauchy \(\Rightarrow a\left(b+1\right)+b=ab+b+a\ge3\sqrt[3]{a^2b^2}\)
\(\Rightarrow P\ge3-\Sigma\dfrac{ab}{\sqrt[3]{a^2b^2}}=3-\Sigma\dfrac{\sqrt[3]{ab}}{3}\)
mà \(\sqrt[3]{ab}=\sqrt[3]{a.b.1}\le\dfrac{a+b+1}{3}\)
\(3-\Sigma\dfrac{\sqrt[3]{ab}}{3}=3-\dfrac{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}{3}\ge3-\dfrac{\dfrac{2\left(a+b+c\right)+3}{3}}{3}=3-1=2\)
\(\Rightarrow P\ge2\) \(\Rightarrow MinP=2\) khi a = b = c =1
Lời giải khác:
Áp dụng BĐT Cauchy-Schwarz:
\(P=\frac{a^2(b+1)}{a+b+ab}+\frac{b^2(c+1)}{b+c+bc}+\frac{c^2(a+1)}{c+a+ac}\)\(=\frac{a^2}{\frac{a+b+ab}{b+1}}+\frac{b^2}{\frac{b+c+bc}{c+1}}+\frac{c^2}{\frac{c+a+ca}{a+1}}\)
\(\geq \frac{(a+b+c)^2}{\frac{(a+1)(b+1)-1}{b+1}+\frac{(b+1)(c+1)-1}{c+1}+\frac{(c+1)(a+1)-1}{a+1}}\)
\(\Leftrightarrow P\geq \frac{9}{a+b+c+3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)}=\frac{9}{6-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq \frac{9}{a+1+b+1+c+1}=\frac{9}{a+b+c+3}=\frac{9}{6}=\frac{3}{2}\)
Do đó: \(6-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\leq 6-\frac{3}{2}=\frac{9}{2}\)
\(\Rightarrow P\geq \frac{9}{\frac{9}{2}}=2\)
Vậy P min là 2
Dấu bằng xảy ra khi \(a=b=c=1\)