Czech and Slovak Mathematical Olympiad

Assume that for some $k, l$, the given inequality holds for all side lengths $a, b, c$ of any triangle. Plugging in $a = c = 1$ and arbitrary $b < 2$, we get $k + lb^2 > 1$. If $k < 1$ then we easily find sufficiently small $b > 0$ that makes the inequality false. Hence $k \ge 1$ and likewise $l \ge 1$.

Let $ABC$ be a triangle in the coordinate plane. Without loss of generality, let $A = [-1, 0]$, $B = [1, 0]$, and denote $C = [x, y]$ ($y \ne 0$). Using Pythagorean Theorem we express the side lengths in terms of $x, y$ and plug it in the given inequality to obtain
$$k((x-1)^2 + y^2) + l((x+1)^2 + y^2) > 4$$
which rewrites as
$$(k+l)x^2 + 2(l-k)x + k + l - 4 > -(k+l)y^2. \quad (1)$$
This inequality has to hold for any $x$ and any $y \ne 0$. However, varying $y$, the right-hand side attains all negative values (recall $k+l > 0$). Therefore for any $x$ we have
$$(k+l)x^2 + 2(l-k)x + (k+l-4) \ge 0, \quad (2)$$
which happens if and only if the discriminant $D = 4(l-k)^2 - 4(k+l-4)(k+l)$ is not positive. The inequality $D \le 0$ rewrites to $kl \ge k+l$.

We found out that if numbers $k, l$ satisfy the given inequality for any triplet of side lengths of a triangle then $k \ge 1, l \ge 1$, and
$$kl \ge k + l. \tag{3}$$
On the other hand, the conjunction of these three conditions is also sufficient: The third condition implies that (2) is satisfied for any real $x$ and since the right-hand side of (1) is negative for $y \ne 0$, inequality (1) is satisfied for any $x$ and any $y \ne 0$. Finally, inequality (1) is equivalent to the given inequality.