Ta có: \(\dfrac{a^3+ab^2}{a^2+b+b^2}=a-\dfrac{ab}{a^2+b+b^2}\ge a-\dfrac{\sqrt[3]{a}}{3}\)
Tương tự:
\(\Rightarrow VT\ge a+b+c-\dfrac{\Sigma\sqrt[3]{a}}{3}=3-\dfrac{\Sigma\sqrt[3]{a}}{3}\)
Áp dụng BĐT cô si chi 3 số dương, ta có:
\(a+1+1\ge3\sqrt[3]{a}\Rightarrow\dfrac{\sqrt[3]{a}}{3}\le\dfrac{a+2}{9}\)
Tương tự:
\(\Rightarrow VT\ge3-\dfrac{a+b+c+6}{9}=3-1=2\left(đpcm\right)\)
Dấu "=" xảy ra <=> a=b=c=1