Why does this method of deriving the classical free particle Lagrangian not work?

Question

I was reading volume two in Landau and Lifshitz's Course of Theoretical Physics (The Classical Theory of Fields). In it, Dr. Landau develops the relativistic Lagrangian as follows: one has $$S=\alpha\int\mathrm{d}s=\int \mathrm{d}t \;L$$ where $\alpha$ is an arbitrary constant. Now, from the Minkowski metric ($\mathrm{d}s^2=c^2\mathrm{d}t^2-\mathrm{d}x^2-\mathrm{d}y^2-\mathrm{d}z^2$) one has $$S=\alpha\int\mathrm{d}s=\alpha\int\mathrm{d}t\,\frac{\mathrm{d}s}{\mathrm{d}t}=\alpha c\int\mathrm{d}t\sqrt{1-\frac{v^2}{c^2}}$$ where one can identify the Lagrangian as $L=\alpha c\sqrt{1-v^2/c^2}$. Expanding this to first order and comparing to the classical case allows one to find $L=-mc^2\sqrt{1-v^2/c^2}$. This is the correct Lagrangian! My question is, why does this not work for normal classical mechanics? We have a metric $\mathrm{d}s^2=\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$, so following similar steps, we have $$S=\alpha\int\mathrm{d}s=\alpha\int\mathrm{d}t\,\frac{\mathrm{d}s}{\mathrm{d}t}=\alpha\int\mathrm{d}t \sqrt{\mathbf{v}\cdot\mathbf{v}}$$ which is clearly incorrect. Why?