Исбот кунед, ки \(x^2+2xy+3y^2+2x+6y+3\geq0\)

\(x^2+2xy+3y^2+2x+6y+3=\)

\(=x^2+2x(y+1)+(y+1)^2-(y+1)^2+3y^2+6y+3=\)

\(=(x+y+1)^2-y^2-2y-1+3y^2+6y+3=\)

\(=(x+y+1)^2+2y^2+4y+2=\)

\(=(x+y+1)^2+2(y^2+2y+1)=\)

\(=(x+y+1)^2+2(y+1)^2\geq0\)

\(x^2+2xy+3y^2+2x+6y+3\geq0\)

Исбот шуд.