\(=\left(2x+y\right)^2+8\left(x^2-\dfrac{1}{2}x+1\right)\)
\(=\left(2x+y\right)^2+8\left(x^2-2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{15}{16}\right)\)
\(=\left(2x+y\right)^2+8\left(x-\dfrac{1}{4}\right)^2+\dfrac{15}{2}\ge\dfrac{15}{2}\forall x\)
Dấu '=' xảy ra khi x=1/4 và y=-1/2