Question

срочноооо!!!!
доказать неравенство: x² + 2xy + 3у³ + 2x + 6y + 4 > 0​

Answer 1

x2+2xy+3y2+2x+6y+4=

=(x2+2xy+y2)+(2y2+6y+2x+4)=

=(x+y)2+2(y2+2y+1)+2x+2y+2=

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

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

=(x+y+1)2+2(y+1)2+1>0