11. Suppose $y(y-1)>x^{2}$, then by $y \geqslant 0$ we know $y>1$. Further, we have $y>\frac{1}{2}+$ $\sqrt{\frac{1}{4}+x^{2}}$. From the assumption $y(y+1) \leqslant(x+1)^{2}$ and $y>1$, we know $y \leqslant-\frac{1}{2}+$ $\sqrt{\frac{1}{4}+(x+1)^{2}}$, thus we get $\frac{1}{2}+\sqrt{\frac{1}{4}+x^{2}}<-\frac{1}{2}+\sqrt{\frac{1}{4}+(x+1)^{2}}$. From this, it is not difficult to deduce $\sqrt{\frac{1}{4}+x^{2}}<x$, a contradiction! Therefore, the original proposition holds.