Ta có:
$\frac{a^3}{b^2+3}=\frac{a^3}{b^2+ab+bc+ca}=\frac{a^3}{(a+b)(b+c)}$
Áp dụng BĐT Cauchy ta có:
$\frac{a^3}{(a+b)(b+c)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3a}{4}$
Hay $\frac{a^3}{b^2+3}\ge\frac{5a}{8}-\frac{b}{4}-\frac{c}{8}$
Tương tự: $\frac{b^3}{c^2+3}\ge\frac{5b}{8}-\frac{c}{4}-\frac{a}{8}$
$\frac{c^3}{a^2+3}\ge\frac{5c}{8}-\frac{a}{4}-\frac{b}{8}$
Từ đó suy ra:
$ \frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\geq \frac{1}{4}(a+b+c)\ge\frac{1}{4}\sqrt{3(ab+bc+ca)}=\frac{3}{4}$
Dấu bằng xảy ra khi: $a=b=c=1$.