Question

доказать неравенство x^2+2xy+3y^2+2x+6y+3>= 0

Answer 1

0≤(х+у+1)²+2(y+1)²=x²+y²+1+2xy+2x+2y+2y²+4y+2=x²+2xy+3y²+2x+6y+3