66th 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 $k+l{b}^2>1$. 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 $$(k+l)x^2+2(l-k)x+k+l-4>-(k+l)y^2.(1)$$ 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 $$(k+l)x^2+2(l-k)x+(k+l-4)\ge 0,(2)$$ 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 $$kl\ge k+l.\text{\hspace{1em}(3)}$$ 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.