Question

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

Answer 1

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

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

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

⇒$(x+3)^{2} +(2y+1)^{2} \geq 0$