Ta có : $a+b=4ab ≤ (a+b)^2$
$\to a+b ≥ 1$
Đặt $P = \dfrac{a}{4b^2+1} + \dfrac{b}{4a^2+1}$
$ = \dfrac{a^2}{4ab^2+a}+\dfrac{b^2}{4a^2b+b}$
$≥ \dfrac{(a+b)^2}{4ab^2+4ab^2+a+b}$ ( Svac - xơ )
$ = \dfrac{(a+b)^2}{(a+b).(4ab+1)} = \dfrac{a+b}{4ab+1}$
$ = \dfrac{a+b}{a+b+1} = 1 - \dfrac{1}{a+b+1} ≥ 1 - \dfrac{1}{1+1} = \dfrac{1}{2}$
Dấu "=" xảy ra $⇔a=b=\dfrac{1}{2}$
Vậy $Min$ $P = \dfrac{1}{2}$ khi $a=b=\dfrac{1}{2}$