(a) Determine the maximum $M$ of $x+y +z$ where $x, y$ and $z$ are positive real numbers with $16xyz = (x + y)^2(x + z)^2$.
(b) Prove the existence of infinitely many triples $(x, y, z)$ of positive rational numbers that satisfy $16xyz = (x + y)^2(x + z)^2$  and $x + y + z = M$.

Proposed by Karl Czakler
### Part (a)
1. **Define Variables:**
   Let \( a = x + y \), \( b = x + z \), and \( c = y + z \). These variables \( a, b, \) and \( c \) must satisfy the triangle inequality, so they can be considered as the sides of a triangle \( \triangle ABC \).

2. **Express \( xyz \) in terms of \( a, b, \) and \( c \):**
   Using the identity for the product of the differences of the sides of a triangle:
   \[
   xyz = \left( \frac{a + b - c}{2} \right) \left( \frac{a - b + c}{2} \right) \left( \frac{-a + b + c}{2} \right)
   \]

3. **Apply Heron's Formula:**
   Heron's formula for the area \( A \) of a triangle with sides \( a, b, \) and \( c \) is:
   \[
   A = \sqrt{ \left( \frac{a + b + c}{2} \right) \left( \frac{a + b - c}{2} \right) \left( \frac{a - b + c}{2} \right) \left( \frac{-a + b + c}{2} \right) }
   \]
   Therefore,
   \[
   \sqrt{ \left( \frac{a + b - c}{2} \right) \left( \frac{a - b + c}{2} \right) \left( \frac{-a + b + c}{2} \right) \left( \frac{a + b + c}{2} \right) } = A
   \]

4. **Relate \( xyz \) to \( A \):**
   Given \( 16xyz = (x + y)^2 (x + z)^2 \), we can rewrite it as:
   \[
   16 \left( \frac{a + b - c}{2} \right) \left( \frac{a - b + c}{2} \right) \left( \frac{-a + b + c}{2} \right) = a^2 b^2
   \]

5. **Simplify the Equation:**
   Using the area \( A \) and the semi-perimeter \( s = \frac{a + b + c}{2} \), we have:
   \[
   a^2 b^2 = \frac{16A^2}{\frac{a + b + c}{2}}
   \]

6. **Express \( A \) in terms of \( \sin \angle ACB \):**
   The area \( A \) can also be written as:
   \[
   A = \frac{ab \cdot \sin(\angle ACB)}{2}
   \]
   Therefore,
   \[
   a + b + c = 8 \cdot \sin^2(\angle ACB)
   \]

7. **Determine the Maximum Value:**
   Since \( 1 \geq \sin^2(\angle ACB) \), it follows that:
   \[
   8 \geq a + b + c \implies 4 \geq x + y + z
   \]

### Part (b)
1. **Equality Condition:**
   Equality holds if \( \sin^2(\angle ACB) = 1 \), which implies \( \angle ACB = 90^\circ \). This means \( c^2 = a^2 + b^2 \).

2. **Substitute \( x = \frac{yz}{4} \):**
   From the equality condition, we have:
   \[
   yz = x^2 + xy + xz = 4x
   \]
   Substituting \( x = \frac{yz}{4} \) into \( x + y + z = 4 \), we get:
   \[
   y = \frac{16 - 4z}{z + 4}
   \]
   and
   \[
   x = \frac{(4 - z)z}{z + 4}
   \]

3. **Rational Solutions:**
   If \( z \) is a positive rational number smaller than 4, then \( x \) and \( y \) must also be rational. This proves the existence of infinitely many triples \( (x, y, z) \) of positive rational numbers that satisfy \( 16xyz = (x + y)^2 (x + z)^2 \) and \( x + y + z = 4 \).

\(\blacksquare\)

The final answer is \( \boxed{ 4 } \)