jmc

Moving all the terms to the left-hand side, we get $$x^2-2xy+2y^2-x+\frac{1}{2}\le 0.$$Multiplying both sides by 2, we get $$2x^2-4xy+4y^2-2x+1\le 0.$$We can write the left-hand side as $$(x^2-4xy+4y^2)+(x^2-2x+1)\le 0,$$which becomes $$(x-2y{)}^2+(x-1{)}^2\le 0.$$By the Trivial Inequality, the only way this can occur is if $x=2y$ and $x=1,$ so $y=\frac{1}{2}.$

So, there is only $\boxed{1}$ ordered pair $(x,y)$ that satisfies the given inequality, namely $(x,y)=\left(1,\frac{1}{2}\right).$