a) Let $n$ be a positive integer. Prove that  $ n\sqrt {x-n^2}\leq \frac {x}{2}$ , for  $x\geq n^2$.
b) Find real $x,y,z$ such that: $ 2\sqrt {x-1} +4\sqrt {y-4} + 6\sqrt {z-9} = x+y+z$
### Part (a)
We need to prove that for a positive integer \( n \) and \( x \geq n^2 \), the inequality \( n\sqrt{x - n^2} \leq \frac{x}{2} \) holds.

1. **Start with the given inequality:**
   \[
   n\sqrt{x - n^2} \leq \frac{x}{2}
   \]

2. **Square both sides to eliminate the square root:**
   \[
   n^2 (x - n^2) \leq \left( \frac{x}{2} \right)^2
   \]

3. **Simplify the left-hand side:**
   \[
   n^2 x - n^4 \leq \frac{x^2}{4}
   \]

4. **Rearrange the inequality:**
   \[
   4n^2 x - 4n^4 \leq x^2
   \]

5. **Bring all terms to one side:**
   \[
   x^2 - 4n^2 x + 4n^4 \geq 0
   \]

6. **Recognize the quadratic form:**
   \[
   (x - 2n^2)^2 \geq 0
   \]

7. **Since the square of any real number is non-negative, the inequality holds:**
   \[
   (x - 2n^2)^2 \geq 0
   \]

Thus, the inequality \( n\sqrt{x - n^2} \leq \frac{x}{2} \) is proven for \( x \geq n^2 \).

\(\blacksquare\)

### Part (b)
We need to find real numbers \( x, y, z \) such that:
\[
2\sqrt{x-1} + 4\sqrt{y-4} + 6\sqrt{z-9} = x + y + z
\]

1. **Apply the result from part (a) to each term:**
   - For \( 2\sqrt{x-1} \leq \frac{x-1}{2} \), we get \( x \geq 1 \).
   - For \( 4\sqrt{y-4} \leq \frac{y-4}{2} \), we get \( y \geq 4 \).
   - For \( 6\sqrt{z-9} \leq \frac{z-9}{2} \), we get \( z \geq 9 \).

2. **Rewrite the inequality using the results from part (a):**
   \[
   2\sqrt{x-1} \leq \frac{x-1}{2}, \quad 4\sqrt{y-4} \leq \frac{y-4}{2}, \quad 6\sqrt{z-9} \leq \frac{z-9}{2}
   \]

3. **Sum the inequalities:**
   \[
   2\sqrt{x-1} + 4\sqrt{y-4} + 6\sqrt{z-9} \leq \frac{x-1}{2} + \frac{y-4}{2} + \frac{z-9}{2}
   \]

4. **Simplify the right-hand side:**
   \[
   \frac{x-1}{2} + \frac{y-4}{2} + \frac{z-9}{2} = \frac{x + y + z - 14}{2}
   \]

5. **Compare with the original equation:**
   \[
   2\sqrt{x-1} + 4\sqrt{y-4} + 6\sqrt{z-9} = x + y + z
   \]

6. **Since the left-hand side is less than or equal to the right-hand side, equality holds only if:**
   \[
   2\sqrt{x-1} = \frac{x-1}{2}, \quad 4\sqrt{y-4} = \frac{y-4}{2}, \quad 6\sqrt{z-9} = \frac{z-9}{2}
   \]

7. **Solve each equality:**
   - \( 2\sqrt{x-1} = \frac{x-1}{2} \) implies \( x = 2 \).
   - \( 4\sqrt{y-4} = \frac{y-4}{2} \) implies \( y = 8 \).
   - \( 6\sqrt{z-9} = \frac{z-9}{2} \) implies \( z = 18 \).

Thus, the solution is \( x = 2 \), \( y = 8 \), and \( z = 18 \).

\(\boxed{(x, y, z) = (2, 8, 18)}\)