Selection Examinations for the IMO

a. First, rewrite the inequality as $x^2-Cx+y^2-Cy+1\ge 0$ and then form perfect squares: $${\left(x-\frac{C}{2}\right)}^2-\frac{C^2}{4}+{\left(y-\frac{C}{2}\right)}^2-\frac{C^2}{4}+1\ge 0.$$ This implies $${\left(x-\frac{C}{2}\right)}^2+{\left(y-\frac{C}{2}\right)}^2+1\ge \frac{C^2}{2}.$$ If $x=y=\frac{C}{2}$, then $1\ge \frac{C^2}{2}$, so $\sqrt{2}\ge C$. The maximum possible real $C$ is equal to $\sqrt{2}$. In this case the inequality holds for all real $x$ and $y$ since it is equivalent to $${\left(x-\frac{\sqrt{2}}{2}\right)}^2+{\left(y-\frac{\sqrt{2}}{2}\right)}^2\ge 0.$$

b. Once again we form perfect squares to get $${\left(x+\frac{y}{2}-\frac{C}{2}\right)}^2-\frac{C^2}{4}+\frac{3}{4}\cdot {\left(y-\frac{C}{3}\right)}^2-\frac{C^2}{12}+1\ge 0.$$ If $y=\frac{C}{3}$ and $x=\frac{C}{2}-\frac{y}{2}=\frac{C}{3}$, then we can multiply by 12 and get $12\ge 3C^2+C^2=4C^2$, so $\sqrt{3}\ge C$. Hence, the maximum possible constant is $C=\sqrt{3}$. In this case the inequality holds for all real $x$ and $y$ since it is equivalent to $${\left(x+\frac{y}{2}-\frac{\sqrt{3}}{2}\right)}^2+\frac{3}{4}\cdot {\left(y-\frac{\sqrt{3}}{3}\right)}^2\ge 0.$$