$Ta^{}$ $có^{}$: (2x-1)$^{2}$ $\geq$ 0 $∀^{}$$x^{}$
/2x-y/ $\geq$ 0 $∀^{}$$x^{}$,$y^{}$
⇒ 3 . /2x-y/ $\geq$ 0 $∀^{}$$x^{}$,$y^{}$
⇒ (2x-1)$^{2}$ + 3 . /2x-y/ + 2017 $\geq$ 2017 $∀^{}$$x^{}$,$y^{}$
Dấu "=" xảy ra khi $\left \{ {{(2x-1)^{2}=0} \atop {3 . /2x-y/=0}} \right.$
⇔ $\left \{ {{2x-1=0} \atop {2x-y=0}} \right.$
⇔ $\left \{ {{2x=1} \atop {y=2x}} \right.$
⇔ $\left \{ {{x=\frac{1}{2}} \atop {y=2.\frac{1}{2}=1}} \right.$
Vậy MinA= 2017 ⇔ $\left \{ {{x= \frac{1}{2}} \atop {y=1}} \right.$
