Which one below is not less than $x^3+y^5$ for all reals $x,y$ such that $0<x<1$ and $0<y<1$?

$ \textbf{(A)}\ x^2y
\qquad\textbf{(B)}\ x^2y^2
\qquad\textbf{(C)}\ x^2y^3
\qquad\textbf{(D)}\ x^3y
\qquad\textbf{(E)}\ xy^4
$
To determine which expression is not less than \(x^3 + y^5\) for all \(0 < x < 1\) and \(0 < y < 1\), we will analyze each option.

1. **Option (A): \(x^2y\)**
   - Let \(x = 2^{-3}\) and \(y = 2^{-2}\).
   - Then \(x^3 = (2^{-3})^3 = 2^{-9}\) and \(y^5 = (2^{-2})^5 = 2^{-10}\).
   - Therefore, \(x^3 + y^5 = 2^{-9} + 2^{-10} < 2^{-8}\).
   - Now, \(x^2y = (2^{-3})^2 \cdot 2^{-2} = 2^{-6} \cdot 2^{-2} = 2^{-8}\).
   - Since \(x^3 + y^5 < 2^{-8} = x^2y\), \(x^2y\) is not less than \(x^3 + y^5\).

2. **Option (B): \(x^2y^2\)**
   - Consider \(x \geq y\):
     - \(x^3 + y^5 > x^3 > x^4 \geq x^2y^2\).
   - Consider \(y > x\):
     - By the Arithmetic Mean-Geometric Mean Inequality (AM-GM), \(x^3 + y^5 \geq 2\sqrt{x^3y^5}\).
     - \(2\sqrt{x^3y^5} > 2\sqrt{x^4y^4} = 2x^2y^2\).
   - Therefore, \(x^3 + y^5 > x^2y^2\).

3. **Option (C): \(x^2y^3\)**
   - \(x^3 + y^5 > x^2y^2 > x^2y^3\) since \(0 < y < 1\).
   - Therefore, \(x^3 + y^5 > x^2y^3\).

4. **Option (D): \(x^3y\)**
   - Consider \(x \geq y\):
     - \(x^3 + y^5 > x^3 > x^4 \geq x^3y\).
   - Consider \(y > x\):
     - By AM-GM, \(x^3 + y^5 \geq 2\sqrt{x^3y^5}\).
     - \(2\sqrt{x^3y^5} > 2\sqrt{x^6y^2} = 2x^3y\).
   - Therefore, \(x^3 + y^5 > x^3y\).

5. **Option (E): \(xy^4\)**
   - Consider \(y \geq x\):
     - \(x^3 + y^5 > y^5 \geq y^4x\).
   - Consider \(x > y\):
     - By AM-GM, \(x^3 + y^5 \geq 2\sqrt{x^3y^5}\).
     - \(2\sqrt{x^3y^5} > 2\sqrt{x^2y^6} = 2xy^3 > xy^4\).
   - Therefore, \(x^3 + y^5 > xy^4\).

Conclusion:
The only option that is not less than \(x^3 + y^5\) for all \(0 < x < 1\) and \(0 < y < 1\) is \(x^2y\).

The final answer is \(\boxed{x^2y}\).