a) Prove that for any nonnegative real $x$,  holds $$x^{\frac32} + 6x^{\frac54} + 8x^{\frac34}\ge 15x.$$ b) Determine all x for which the equality holds
1. **Substitution**: Let \( x = Y^4 \). Then, we have:
   \[
   x^{\frac{3}{2}} = (Y^4)^{\frac{3}{2}} = Y^6, \quad x^{\frac{5}{4}} = (Y^4)^{\frac{5}{4}} = Y^5, \quad x^{\frac{3}{4}} = (Y^4)^{\frac{3}{4}} = Y^3
   \]
   Substituting these into the original inequality, we get:
   \[
   Y^6 + 6Y^5 + 8Y^3 \ge 15Y^4
   \]

2. **Rearranging the inequality**: Move all terms to one side of the inequality:
   \[
   Y^6 + 6Y^5 + 8Y^3 - 15Y^4 \ge 0
   \]

3. **Factoring**: Factor out the common term \( Y^3 \):
   \[
   Y^3(Y^3 + 6Y^2 - 15Y + 8) \ge 0
   \]

4. **Further factoring**: Notice that \( Y^3 + 6Y^2 - 15Y + 8 \) can be factored as:
   \[
   Y^3(Y - 1)^2(Y + 8) \ge 0
   \]

5. **Analyzing the factors**: Since \( Y^3 \ge 0 \) for \( Y \ge 0 \), \( (Y - 1)^2 \ge 0 \) for all \( Y \), and \( Y + 8 \ge 0 \) for all \( Y \ge -8 \), the product \( Y^3(Y - 1)^2(Y + 8) \ge 0 \) holds for all \( Y \ge 0 \).

6. **Equality condition**: The equality holds when:
   \[
   Y^3(Y - 1)^2(Y + 8) = 0
   \]
   This occurs when \( Y = 0 \) or \( Y = 1 \). 

7. **Back-substitution**: Recall that \( x = Y^4 \). Therefore:
   \[
   \text{If } Y = 0, \text{ then } x = 0^4 = 0
   \]
   \[
   \text{If } Y = 1, \text{ then } x = 1^4 = 1
   \]

Thus, the equality holds for \( x = 0 \) and \( x = 1 \).

The final answer is \( \boxed{ x = 0 } \) and \( x = 1 \).