Find all pairs of real numbers $k, l$ such that inequality $ka^2 + lb^2> c^2$ applies to the lengths of sides $a, b, c$ of any triangle.
1. **Applying the Law of Cosines:**
   The given inequality is \( ka^2 + lb^2 > c^2 \). Using the Law of Cosines, we have:
   \[
   c^2 = a^2 + b^2 - 2ab \cos{\angle{C}}
   \]
   Substituting this into the inequality, we get:
   \[
   ka^2 + lb^2 > a^2 + b^2 - 2ab \cos{\angle{C}}
   \]
   Simplifying, we obtain:
   \[
   (k-1)a^2 + (l-1)b^2 > -2ab \cos{\angle{C}}
   \]

2. **Considering the Range of \(\cos{\angle{C}}\):**
   Since \(\cos{\angle{C}}\) can take values arbitrarily close to \(-1\) (but not exactly \(-1\) because \(A\), \(B\), and \(C\) would be collinear), we must have:
   \[
   (k-1)a^2 + (l-1)b^2 \geq 2ab
   \]

3. **Analyzing the Case When One of \(k\) or \(l\) is Not Larger Than 1:**
   Suppose \(k \leq 1\). Then:
   \[
   (k-1)a^2 + (l-1)b^2 \leq (l-1)b^2
   \]
   This implies:
   \[
   (l-1)b^2 \geq 2ab
   \]
   This inequality cannot hold for sufficiently large \(a\). Therefore, both \(k\) and \(l\) must be greater than 1.

4. **Ensuring the Inequality Holds:**
   We need to ensure:
   \[
   (k-1)a^2 + (l-1)b^2 \geq 2ab \sqrt{(k-1)(l-1)}
   \]
   If \((k-1)(l-1) \geq 1\), then:
   \[
   (k-1)a^2 + (l-1)b^2 \geq 2ab
   \]
   This condition is necessary for the initial inequality to hold.

5. **Showing the Necessity of the Condition:**
   If \((k-1)(l-1) < 1\), then:
   \[
   (k-1)a^2 - 2ab + (l-1)b^2 \geq 0
   \]
   This is a quadratic inequality in \(a\) with a non-positive discriminant:
   \[
   D = (2b)^2 - 4(k-1)(l-1)b^2 = 4b^2(1 - (k-1)(l-1))
   \]
   If \((k-1)(l-1) < 1\), then \(D > 0\), which is a contradiction. Therefore, \((k-1)(l-1) \geq 1\) is necessary.

6. **Conclusion:**
   The suitable pairs \((k, l)\) are those that satisfy:
   \[
   (k-1)(l-1) \geq 1 \quad \text{and} \quad k, l > 1
   \]
   This can be rewritten as:
   \[
   kl \geq k + l \quad \text{and} \quad k, l > 1
   \]

The final answer is \( \boxed{ (k, l) } \) such that \(kl \geq k + l\) and \(k, l > 1\).