Áp dụng $BĐT$ $AM-GM$ ta có :
$\dfrac{1}{a^4+b^2+2ab^2}+\dfrac{1}{b^4+a^2+2ab^2}$
$≤ \dfrac{1}{2a^2b+2ab^2}+\dfrac{1}{2b^2a+2a^2b}$
$ = \dfrac{2}{2.ab.(a+b)}$
$ = \dfrac{1}{ab.(a+b)}$
$≤ \dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{a+b}$
Ta thấy " $\dfrac{1}{a}+\dfrac{1}{b} ≥ \dfrac{4}{a+b}$
$\to 2 ≥ \dfrac{4}{a+b}$
$\to \dfrac{1}{2} ≥ \dfrac{1}{a+b}$
Lại có $\dfrac{1}{a}.\dfrac{1}{b} ≤ \dfrac{\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)^2}{4} = 1$
Do đó : $Q ≤ \dfrac{1}{2}.1=\dfrac{1}{2}$
Dấu "=" xảy ra $⇔a=b=1$
Vậy $Q_{min} = \dfrac{1}{2}$ tại $a=b=1$
