Ta có $\frac{a^{3}}{(b+c)^{2}}+\frac{a}{4}\geq 2\sqrt{\frac{a^{4}}{4(b+c)^{2}}}=\frac{a^{2}}{b+c}$
Tương tự
$\Rightarrow \frac{a^{3}}{(b+c)^{2}}+\frac{b^{3}}{(a+c)^{2}}+\frac{c^{3}}{(a+b)^{2}}+\frac{a+b+c}{4}\geq \frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}$
$\geq \frac{(a+b+c)^{2}}{2(a+b+c)}=1/2$
$\Rightarrow \frac{a^{3}}{(b+c)^{2}}+\frac{b^{3}}{(a+c)^{2}}+\frac{c^{3}}{(a+b)^{2}}\geq \frac{1}{2}-\frac{(a+b+c)}{4}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\Rightarrow đpcm$
Dấu bằng xảy ra $\Leftrightarrow$a=b=c=1/3