Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$

Or alternatively, by AM-GM we get $${x^2+y^2}\geq 2|xy|$$ so $$\frac{x^2y^2}{x^2+y^2}\le \frac{x^2y^2}{2|xy|}=\frac{1}{2}|xy|$$ and this tends to zero if $x,y$ tend to zero.