Tu gia thiet suy ra $12\geq \left| {2y-x} \right|+\left| {4x+y} \right|\geq \left| {2y-x+4x+y} \right|$
$<=>12\geq3 \left| {x+y} \right|$
$<=>\left| {x+y} \right|\leq 4$
$<=>(x+y)^{2}\leq 16$
Co $xy\leq \frac{(x+y)^{2}}{4}\leq 4$
Dau bang khi $x=y=\pm 2$