Let a,b,c and d be non-negative real number such that `a ^(5)+b^(5) le 1 and c ^(5) + d^(5) le 1.` Find the maximum value of `a ^(2) c ^(3) +b^(2) d ^(3).`

Answer

To solve the problem, we need to find the maximum value of the expression \( a^2 c^3 + b^2 d^3 \) given the constraints \( a^5 + b^5 \leq 1 \) and \( c^5 + d^5 \leq 1 \). We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to derive the solution.

### Step-by-step Solution:

1. **Express the terms in a suitable form**:
We can rewrite \( a^2 c^3 \) and \( b^2 d^3 \) using the properties of exponents:
\[ a^2 c^3 = a^2 c^3 = (a^5)^{2/5} (c^5)^{3/5} \]
\[ b^2 d^3 = b^2 d^3 = (b^5)^{2/5} (d^5)^{3/5} \]

2. **Apply AM-GM inequality**:
According to the AM-GM inequality, for non-negative numbers \( x_1, x_2, \ldots, x_n \):
\[ \frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n} \]
We apply this to \( a^5, a^5, c^5, c^5, c^5 \) (5 terms):
\[ \frac{a^5 + a^5 + c^5 + c^5 + c^5}{5} \geq (a^5)^2 (c^5)^3)^{1/5} \]
Simplifying gives:
\[ \frac{2a^5 + 3c^5}{5} \geq a^2 c^3 \]

3. **Similarly for \( b^2 d^3 \)**:
Using the same approach for \( b^5, b^5, d^5, d^5, d^5 \):
\[ \frac{2b^5 + 3d^5}{5} \geq b^2 d^3 \]

4. **Combine the inequalities**:
Adding the two inequalities:
\[ \frac{2a^5 + 3c^5}{5} + \frac{2b^5 + 3d^5}{5} \geq a^2 c^3 + b^2 d^3 \]
This simplifies to:
\[ \frac{2(a^5 + b^5) + 3(c^5 + d^5)}{5} \geq a^2 c^3 + b^2 d^3 \]

5. **Use the constraints**:
Since \( a^5 + b^5 \leq 1 \) and \( c^5 + d^5 \leq 1 \), we can substitute:
\[ \frac{2(1) + 3(1)}{5} \geq a^2 c^3 + b^2 d^3 \]
This gives:
\[ \frac{5}{5} \geq a^2 c^3 + b^2 d^3 \]
Hence:
\[ a^2 c^3 + b^2 d^3 \leq 1 \]

6. **Conclusion**:
The maximum value of \( a^2 c^3 + b^2 d^3 \) is therefore:
\[ \boxed{1} \]