If $ x,y$ are positive real numbers with sum $ 2a$, prove that :


$ x^3y^3(x^2\plus{}y^2)^2 \leq 4a^{10}$

When does equality hold ? 

Babis
1. Let \( x + y = 2a \) and \( xy = b \). By the AM-GM inequality, we have:
   \[
   x + y \geq 2\sqrt{xy} \implies 2a \geq 2\sqrt{b} \implies a^2 \geq b
   \]
2. We need to prove that:
   \[
   x^3 y^3 (x^2 + y^2)^2 \leq 4a^{10}
   \]
3. Express \( x^3 y^3 \) and \( x^2 + y^2 \) in terms of \( a \) and \( b \):
   \[
   x^3 y^3 = b^3
   \]
   \[
   x^2 + y^2 = (x + y)^2 - 2xy = 4a^2 - 2b
   \]
4. Substitute these into the inequality:
   \[
   b^3 (4a^2 - 2b)^2 \leq 4a^{10}
   \]
5. Expand and simplify the left-hand side:
   \[
   b^3 (4a^2 - 2b)^2 = b^3 (16a^4 - 16a^2 b + 4b^2)
   \]
   \[
   = 16a^4 b^3 - 16a^2 b^4 + 4b^5
   \]
6. We need to show:
   \[
   16a^4 b^3 - 16a^2 b^4 + 4b^5 \leq 4a^{10}
   \]
   Divide both sides by 4:
   \[
   4a^4 b^3 - 4a^2 b^4 + b^5 \leq a^{10}
   \]
7. Rearrange the inequality:
   \[
   a^{10} \geq 4a^4 b^3 - 4a^2 b^4 + b^5
   \]
8. By the AM-GM inequality, we have:
   \[
   a^{10} + 3a^2 b^4 \geq 4a^4 b^3 \quad \text{(1)}
   \]
   \[
   a^2 b^4 \geq b^5 \quad \text{(2)}
   \]
9. Combining (1) and (2):
   \[
   a^{10} + 3a^2 b^4 \geq 4a^4 b^3
   \]
   \[
   a^2 b^4 \geq b^5
   \]
   Adding these inequalities:
   \[
   a^{10} + 4a^2 b^4 \geq 4a^4 b^3 + b^5
   \]
   This completes the proof.

Equality holds when \( x = y \). In this case, \( x = y = a \), and we have:
\[
x^3 y^3 (x^2 + y^2)^2 = a^3 a^3 (a^2 + a^2)^2 = a^6 (2a^2)^2 = a^6 \cdot 4a^4 = 4a^{10}
\]

The final answer is \( \boxed{ 4a^{10} } \).