Áp dụng BĐT `Cauchy`
`A=16(x^4+y^4)+1/(4xy)`
`A\ge 16.2.\sqrt[x^4.y^4]+1/(4xy)`
`A\ge 32x^2y^2+1/(8xy)+1/(8xy)`
`A\ge 3\root{3}{32x^2y^2. 1/(8xy). 1/(8xy)}`
`A\ge 3/\root{3}{2}`
Dấu `=` xảy ra $⇔\begin{cases}x=y\\32x^2y^2=\dfrac{1}{8xy}\end{cases}⇔x=y=±\dfrac{1}{\sqrt[3]{16}}$
Vậy $Min_A=\dfrac{3}{\sqrt[3]{2}}⇔x=y=±\dfrac{1}{\sqrt[3]{16}}$