jmc

The given inequality expands as $$x^2+y^2+1\ge Cx+Cy.$$Completing the square in $x$ and $y,$ we get $${\left(x-\frac{C}{2}\right)}^2+{\left(y-\frac{C}{2}\right)}^2+1-\frac{C^2}{2}\ge 0.$$This inequality holds for all $x$ and $y$ if and only if $1-\frac{C^2}{2}\ge 0,$ or $C^2\le 2.$ Thus, the largest possible value of $C$ is $\boxed{\sqrt{2}}.$