Áp dụng BĐT Cauchy ta có:
$\dfrac{a^2}{\sqrt{b+3}}+\dfrac{\sqrt{b+3}}{4}\ge2\sqrt{\dfrac{a^2}{\sqrt{b+3}}.\dfrac{\sqrt{b+3}}{4}}=a$
$\Rightarrow \dfrac{a^2}{\sqrt{b+3}}\ge a-\dfrac{\sqrt{b+3}}{4}\ge a-\dfrac{1}{16}[(b+3)+4]=a-\dfrac{b}{16}-\dfrac{7}{16}$
Tương tự: $\dfrac{b^2}{\sqrt{c+3}}\ge b-\dfrac{c}{16}-\dfrac{7}{16}$
$\dfrac{c^2}{\sqrt{a+3}}\ge c-\dfrac{a}{16}-\dfrac{7}{16}$
Cộng các BĐT trên, ta được:
$\dfrac{a^2}{\sqrt{b+3}}+\dfrac{b^2}{\sqrt{c+3}}+\dfrac{c^2}{\sqrt{a+3}}\ge \dfrac{15(a+b+c)}{16}-\dfrac{21}{16}=\dfrac{3}{2}$
Dấu bằng xảy ra khi $a=b=c=1$.