ta có $\frac{a}{bc(1+a^2)} = \frac{a}{bc+a(a+b+c)}= \frac{a}{(a+b)(a+c)}\leq \frac{a}{2\sqrt{ab}2\sqrt{ac}}= \frac{1}{4\sqrt{bc}}$
Tương tự ta có
$S = \frac{a}{bc(1+a^2)}+\frac{b}{ac(1+b^2)}+\frac{c}{ab(1+c^2)} \leq \frac{1}{4\sqrt{bc}} +\frac{1}{4\sqrt{ac}}+\frac{1}{4\sqrt{ab}} = \frac{\sqrt a+\sqrt b+\sqrt c}{4\sqrt{abc}}$
Theo bất đẳng thức bunhia
$(\sqrt a +\sqrt b+ \sqrt c)=\sqrt{(1.\sqrt a +1.\sqrt b+ 1.\sqrt c)^2}\leq \sqrt{(1+1+1)(a+b+c)}=\sqrt {3(a+b+c)}=\sqrt 3\sqrt{abc}$
$S\leq \frac{\sqrt 3\sqrt{abc}}{4\sqrt{abc}} = \frac{\sqrt 3}{4}$
Vậy max $S = \frac{\sqrt 3}{4}$
Dấu bằng khi $a=b=c = \sqrt 3$
Nhớ vote nhé