Let $r_1,r_2,\dots ,r_n$ be real numbers. Given $n$ reals $a_1,a_2,\dots ,a_n$ that are not all equal to $0$, suppose that inequality
\[r_1(x_1-a_1)+ r_2(x_2-a_2)+\dots + r_n(x_n-a_n)\leq\sqrt{x_1^2+ x_2^2+\dots + x_n^2}-\sqrt{a_1^2+a_2^2+\dots +a_n^2}\]
holds for arbitrary reals $x_1,x_2,\dots ,x_n$. Find the values of $r_1,r_2,\dots ,r_n$.
1. **Interpret the given inequality:**
   \[
   r_1(x_1 - a_1) + r_2(x_2 - a_2) + \dots + r_n(x_n - a_n) \leq \sqrt{x_1^2 + x_2^2 + \dots + x_n^2} - \sqrt{a_1^2 + a_2^2 + \dots + a_n^2}
   \]
   This inequality must hold for all real numbers \(x_1, x_2, \dots, x_n\).

2. **Rewriting the inequality in vector form:**
   Let \(\mathbf{r} = (r_1, r_2, \dots, r_n)\), \(\mathbf{x} = (x_1, x_2, \dots, x_n)\), and \(\mathbf{a} = (a_1, a_2, \dots, a_n)\). The inequality can be rewritten as:
   \[
   \mathbf{r} \cdot (\mathbf{x} - \mathbf{a}) \leq \|\mathbf{x}\| - \|\mathbf{a}\|
   \]
   where \(\mathbf{r} \cdot (\mathbf{x} - \mathbf{a})\) denotes the dot product and \(\|\mathbf{x}\|\) denotes the Euclidean norm.

3. **Analyzing the function:**
   Consider the function \(f(\mathbf{x}) = \mathbf{r} \cdot \mathbf{x} - \|\mathbf{x}\|\). The given inequality implies that \(\mathbf{a}\) is a global minimum of this function.

4. **Finding the gradient:**
   The gradient of \(f(\mathbf{x})\) is:
   \[
   \nabla f(\mathbf{x}) = \mathbf{r} - \frac{\mathbf{x}}{\|\mathbf{x}\|}
   \]
   At the point \(\mathbf{x} = \mathbf{a}\), since \(\mathbf{a}\) is a global minimum, the gradient must be zero:
   \[
   \nabla f(\mathbf{a}) = \mathbf{r} - \frac{\mathbf{a}}{\|\mathbf{a}\|} = 0
   \]
   Therefore,
   \[
   \mathbf{r} = \frac{\mathbf{a}}{\|\mathbf{a}\|}
   \]

5. **Verifying the condition:**
   Substitute \(\mathbf{r} = \frac{\mathbf{a}}{\|\mathbf{a}\|}\) back into the original inequality:
   \[
   \frac{a_1}{\|\mathbf{a}\|}(x_1 - a_1) + \frac{a_2}{\|\mathbf{a}\|}(x_2 - a_2) + \dots + \frac{a_n}{\|\mathbf{a}\|}(x_n - a_n) \leq \sqrt{x_1^2 + x_2^2 + \dots + x_n^2} - \sqrt{a_1^2 + a_2^2 + \dots + a_n^2}
   \]
   Simplifying the left-hand side:
   \[
   \frac{1}{\|\mathbf{a}\|} \left( a_1(x_1 - a_1) + a_2(x_2 - a_2) + \dots + a_n(x_n - a_n) \right)
   \]
   This can be written as:
   \[
   \frac{1}{\|\mathbf{a}\|} \left( \mathbf{a} \cdot \mathbf{x} - \|\mathbf{a}\|^2 \right)
   \]
   The inequality becomes:
   \[
   \frac{\mathbf{a} \cdot \mathbf{x} - \|\mathbf{a}\|^2}{\|\mathbf{a}\|} \leq \|\mathbf{x}\| - \|\mathbf{a}\|
   \]
   Multiplying both sides by \(\|\mathbf{a}\|\):
   \[
   \mathbf{a} \cdot \mathbf{x} - \|\mathbf{a}\|^2 \leq \|\mathbf{a}\| \|\mathbf{x}\| - \|\mathbf{a}\|^2
   \]
   Simplifying:
   \[
   \mathbf{a} \cdot \mathbf{x} \leq \|\mathbf{a}\| \|\mathbf{x}\|
   \]
   This is the Cauchy-Schwarz inequality, which holds true for all vectors \(\mathbf{a}\) and \(\mathbf{x}\).

Conclusion:
\[
\boxed{r_i = \frac{a_i}{\|\mathbf{a}\|} \text{ for } i = 1, 2, \dots, n}
\]