$k,l>1, k+l>kl$ Ravi substitution and limits at zero.

Applying the law of cosines, the given condition becomes: $$ka^2 + lb^2> a^2+b^2-2ab \cdot \cos{\angle{C}} \Longleftrightarrow$$$$(k-1)a^2 + (l-1)b^2> -2ab \cdot \cos{\angle{C}} \Longleftrightarrow$$Since $\cos{\angle{C}}$ can take values arbitrarily close to $-1$ (but not $-1$, since $A$, $B$, $C$ would be collinear) we must have: $$(k-1)a^2 + (l-1)b^2 \geq 2ab $$If one of $k$, $l$ is not larger than $1$, WLOG $k \leq 1$, then $$(k-1)a^2 + (l-1)b^2 \leq (l-1)b^2 \Longrightarrow$$$$(l-1)b^2 \geq 2ab$$which cannot hold for sufficiently large $a$. So $k$, $l$$>1$. Note that: $$(k-1)a^2 + (l-1)b^2 \geq 2ab \cdot \sqrt{(k-1)(l-1)}$$If $(k-1)(l-1) \geq 1$, we are done. We will show that this condition is necessary for the initial condition to be true. If $(k-1)(l-1) < 1$, then we have: $$(k-1)a^2 - 2ab + (l-1)b^2 \geq 0$$This is a trinomial with variable $a$ and has a non-positive discriminant: $$D=(2b)^2-4(k-1)(l-1)b^2 \Longleftrightarrow$$$$D=4b^2(1-(k-1)(l-1))>0$$which is a contradiction. In conclusion, the suitable pairs $(k,l)$ are those who satisfy $(k-1)(l-1) \geq 1\Longleftrightarrow kl \geq k+l$ and $k$, $l$$>1$.