Determine all pairs $(k,n)$ of non-negative integers such that the following inequality holds $\forall x,y>0$:
\[1+ \frac{y^n}{x^k} \geq \frac{(1+y)^n}{(1+x)^k}.\]
1. **Initial Assumptions and Simplification:**
   We need to determine all pairs \((k, n)\) of non-negative integers such that the inequality 
   \[
   1 + \frac{y^n}{x^k} \geq \frac{(1+y)^n}{(1+x)^k}
   \]
   holds for all \(x, y > 0\). 

2. **Special Cases:**
   - If \(n = 0\), the inequality simplifies to:
     \[
     1 + \frac{y^0}{x^k} = 1 + \frac{1}{x^k} \geq \frac{(1+y)^0}{(1+x)^k} = \frac{1}{(1+x)^k}
     \]
     This is always true for any \(k \geq 0\) because \(1 + \frac{1}{x^k} \geq \frac{1}{(1+x)^k}\).

   - If \(k = 0\), the inequality simplifies to:
     \[
     1 + \frac{y^n}{x^0} = 1 + y^n \geq \frac{(1+y)^n}{(1+x)^0} = (1+y)^n
     \]
     This is always true for any \(n \geq 0\) because \(1 + y^n \geq (1+y)^n\).

3. **General Case \(n, k \geq 1\):**
   - We need to show that \(k \geq n-1\) is necessary. By taking \(x = y\), the inequality becomes:
     \[
     1 + x^{n-k} \geq \frac{(1+x)^n}{(1+x)^k} = (1+x)^{n-k}
     \]
     This inequality holds if and only if \(n - k \leq 1\), which implies \(k \geq n - 1\).

4. **Verification for \(k \geq n-1\):**
   - For \(k \geq n\), consider two cases:
     - If \(x \geq y\), then \(\frac{y^n}{x^k} \leq 1\) and the inequality \(1 + \frac{y^n}{x^k} \geq 1\) is trivially true.
     - If \(y > x\), then \(\frac{y^n}{x^k} \geq \frac{(1+y)^n}{(1+x)^k}\) because \(k \geq n\) ensures that the growth rate of the numerator is higher than the denominator.

   - For \(k = n-1\), we need to show:
     \[
     1 + \frac{y^n}{x^{n-1}} \geq \frac{(1+y)^n}{(1+x)^{n-1}}
     \]
     This can be shown using Radon's inequality, which states that for non-negative \(a, b, c, d\):
     \[
     \frac{a}{b} + \frac{c}{d} \geq \frac{a+c}{b+d}
     \]
     Applying this inequality appropriately confirms the given inequality.

Conclusion:
The pairs \((k, n)\) that satisfy the inequality are:
- \((0, k)\) for any \(k \geq 0\)
- \((1, 0)\)
- \((0, 0)\)
- \((k, n)\) where \(k \geq n-1 \geq 1\)

The final answer is \( \boxed{ (0, k) } \) for any \(k \geq 0\), \((1, 0)\), \((0, 0)\), and \((k, n)\) where \(k \geq n-1 \geq 1\).