If $\zeta$ denote the Riemann Zeta Function, and $s>1$ then $$\sum \limits_{k=1}^{\infty} \frac{1}{1+k^s}\geq \frac{\zeta (s)}{1+\zeta (s)}$$
To prove the inequality 
\[ \sum_{k=1}^{\infty} \frac{1}{1+k^s} \geq \frac{\zeta(s)}{1+\zeta(s)}, \]
we will use the following lemma:

**Lemma:** For any non-negative real numbers \( x_1, x_2, \ldots, x_n \), the following inequality holds:
\[ \sum_{k=1}^n \frac{x_k}{1+x_k} \geq \frac{\sum_{k=1}^n x_k}{1+\sum_{k=1}^n x_k}. \]

**Proof of Lemma:**
1. **Base Case:**
   For \( n = 1 \), the inequality is trivially true:
   \[ \frac{x_1}{1+x_1} \geq \frac{x_1}{1+x_1}. \]

2. **Inductive Step:**
   Assume the inequality holds for some \( n \), i.e.,
   \[ \sum_{k=1}^n \frac{x_k}{1+x_k} \geq \frac{\sum_{k=1}^n x_k}{1+\sum_{k=1}^n x_k}. \]
   
   We need to show that it holds for \( n+1 \):
   \[ \sum_{k=1}^{n+1} \frac{x_k}{1+x_k} \geq \frac{\sum_{k=1}^{n+1} x_k}{1+\sum_{k=1}^{n+1} x_k}. \]
   
   Consider the left-hand side:
   \[ \sum_{k=1}^{n+1} \frac{x_k}{1+x_k} = \sum_{k=1}^n \frac{x_k}{1+x_k} + \frac{x_{n+1}}{1+x_{n+1}}. \]
   
   By the inductive hypothesis, we have:
   \[ \sum_{k=1}^n \frac{x_k}{1+x_k} \geq \frac{\sum_{k=1}^n x_k}{1+\sum_{k=1}^n x_k}. \]
   
   Let \( S_n = \sum_{k=1}^n x_k \). Then:
   \[ \sum_{k=1}^n \frac{x_k}{1+x_k} + \frac{x_{n+1}}{1+x_{n+1}} \geq \frac{S_n}{1+S_n} + \frac{x_{n+1}}{1+x_{n+1}}. \]
   
   We need to show:
   \[ \frac{S_n}{1+S_n} + \frac{x_{n+1}}{1+x_{n+1}} \geq \frac{S_n + x_{n+1}}{1 + S_n + x_{n+1}}. \]
   
   Consider the function \( f(x) = \frac{x}{1+x} \). This function is concave, as its second derivative \( f''(x) = -\frac{2}{(1+x)^3} \) is negative. By Jensen's inequality for concave functions, we have:
   \[ f\left( \frac{S_n + x_{n+1}}{2} \right) \geq \frac{f(S_n) + f(x_{n+1})}{2}. \]
   
   Simplifying, we get:
   \[ \frac{S_n + x_{n+1}}{2(1 + \frac{S_n + x_{n+1}}{2})} \geq \frac{\frac{S_n}{1+S_n} + \frac{x_{n+1}}{1+x_{n+1}}}{2}. \]
   
   Multiplying both sides by 2, we obtain:
   \[ \frac{S_n + x_{n+1}}{2 + S_n + x_{n+1}} \geq \frac{S_n}{1+S_n} + \frac{x_{n+1}}{1+x_{n+1}}. \]
   
   This completes the inductive step, proving the lemma.

3. **Application of Lemma:**
   Let \( x_k = \frac{1}{k^s} \). Then, applying the lemma, we get:
   \[ \sum_{k=1}^n \frac{\frac{1}{k^s}}{1+\frac{1}{k^s}} \geq \frac{\sum_{k=1}^n \frac{1}{k^s}}{1+\sum_{k=1}^n \frac{1}{k^s}}. \]
   
   Simplifying the left-hand side:
   \[ \sum_{k=1}^n \frac{\frac{1}{k^s}}{1+\frac{1}{k^s}} = \sum_{k=1}^n \frac{1}{k^s + 1}. \]
   
   As \( n \to \infty \), the right-hand side becomes:
   \[ \frac{\sum_{k=1}^\infty \frac{1}{k^s}}{1+\sum_{k=1}^\infty \frac{1}{k^s}} = \frac{\zeta(s)}{1+\zeta(s)}. \]
   
   Therefore, we have:
   \[ \sum_{k=1}^\infty \frac{1}{1+k^s} \geq \frac{\zeta(s)}{1+\zeta(s)}. \]

\(\blacksquare\)

The final answer is \( \boxed{ \sum_{k=1}^{\infty} \frac{1}{1+k^s} \geq \frac{\zeta(s)}{1+\zeta(s)} } \)