2016
Problem - 3970
Find all real numbers $m$ such that $$x^2+my^2-4my+6y-6x+2m+8 \ge 0$$ for every pair of real numbers $x$ and $y$.
Rewrite the equation with respect to $x$: $$x^2 -6x +my^2-4my+6y+2m+8\ge 0\implies (-6)^2 -4(my^2-4my+6y+2m+8)\le 0 $$
Rewrite the above relationship with respect to $y$: $$-my^2 +(4m - 6)y +(1-2m)\le 0 $$
This implies $m>0$ and $$(4m-6)^2 +4m(1-2m)\le 0\implies 1\le m \le\frac{9}{2}$$