Question

x^2+4y^2+6x+4y+10≥0
Доказать неровномерность

Answer 1

$x^{2} +4y^{2} +6x+4y+10geq 0\ x^{2} +6x+9+4y^{2} +4y+1geq 0\ (x+3)^{2} +(2y+1)^{2} geq 0$

$(x+3)^{2} geq 0$

$(2y+1)^{2}geq 0$

⇒$(x+3)^{2} +(2y+1)^{2} geq 0$