Để ý rằng : $b^3+b^2+4=b^3+2b^2-b^2-2b+2b+4$
$ = (b+2).(b^2-b+2)$
Do $a,b,c>0$ nên theo BĐT Cô - si ta có :
$0<\sqrt[]{b^3+b^2+4} = \sqrt[]{(b+2).(b^2-b+2)} ≤ \dfrac{b+2+b^2-b+2}{2} = \dfrac{b^2+4}{2}$
$⇒ \dfrac{a}{\sqrt[]{b^3+b^2+4}} ≥ \dfrac{2a}{b^2+4}$
Lại có : $\dfrac{2a}{b^2+4b} = \dfrac{a}{2}.\dfrac{4}{b^2+4}$
$ = \dfrac{a}{2}.\dfrac{b^2+4-b^2}{b^2+4} = \dfrac{a}{2}.\bigg(1-\dfrac{b^2}{b^2+4}\bigg)$
$≥ \dfrac{a}{2}.\bigg(1-\dfrac{b^2}{4b}\bigg) = \dfrac{a}{2} - \dfrac{ab}{8}$
Vì vậy $\dfrac{a}{\sqrt[]{b^3+b^2+4}} ≥ \dfrac{a}{2}-\dfrac{ab}{8}$
Tương tự ta cũng có : $\dfrac{b}{\sqrt[]{c^3+c^2+4}} ≥ \dfrac{b}{2}-\dfrac{bc}{8}$
$\dfrac{c}{\sqrt[]{a^3+a^2+4}} ≥ \dfrac{c}{2}-\dfrac{ca}{8}$
Do đó : $\dfrac{a}{\sqrt[]{b^3+b^2+4}} + \dfrac{b}{\sqrt[]{c^3+c^2+4}} + \dfrac{c}{\sqrt[]{a^3+a^2+4}} ≥ \dfrac{a+b+c}{2}-\dfrac{ab+bc+ca}{4} = 3- \dfrac{ab+bc+ca}{8}$
Áp dụng BĐT $xy+yz+zx ≤ \dfrac{(x+y+z)^2}{3}$ ta có :
$ab+bc+ca ≤ \dfrac{(a+b+c)^2}{3} = \dfrac{6^2}{3} = 12$
$\to 3 - \dfrac{ab+bc+ca}{8} ≥3- \dfrac{3}{2} = \dfrac{3}{2}$
Hay : $\dfrac{a}{\sqrt[]{b^3+b^2+4}} + \dfrac{b}{\sqrt[]{c^3+c^2+4}} + \dfrac{c}{\sqrt[]{a^3+a^2+4}} ≥ \dfrac{3}{2}$
Dấu "=" xảy ra $⇔a=b=c=2$
