2. Find all pairs of real numbers $k, l$ such that the inequality

$$
k a^{2}+l b^{2}>c^{2}
$$

holds for the side lengths $a, b, c$ of any triangle.

(Patrik Bak)
Solution. Suppose that the given inequality holds for some pair $k, l$ for the side lengths $a, b, c$ of any triangle. If we substitute $a=1, c=1$ and any positive $b>1$ into it, we get an inequality that must hold. If $k<1$, for sufficiently small positive $b$, this inequality would no longer hold, so we must have $k \geq 1$. Similarly, we must have $l \geq 1$.

In an orthonormal coordinate system, let us choose points $A[-1,0], B[1,0], C[x, y]$; points $A, B, C$ are vertices of a triangle for any real $x$ and any real $y \neq 0$, and up to similarity, any triangle can be placed in this way. By substituting the side lengths of triangle $ABC$ (which can be easily calculated using the Pythagorean theorem) into the given inequality, it transforms into the form

$$
k\left((x-1)^{2}+y^{2}\right)+l\left((x+1)^{2}+y^{2}\right)>4
$$

or equivalently,

$$
(k+l) x^{2}+2(l-k) x+k+l-4>-(k+l) y^{2}.
$$

This must hold for every $x$ and any $y \neq 0$. For a fixed $x$, we can achieve any negative value of the expression $V(y)=-(k+l) y^{2}$ on the right side of the above inequality by choosing the value of $y$, since (as we already know) $k+l>0$. Therefore, for every $x$, we must have

$$
(k+l) x^{2}+2(l-k) x+(k+l-4) \geq 0,
$$

which, given the positive coefficient of $x^{2}$ on the left side, holds if and only if the corresponding discriminant $D=4(l-k)^{2}-4(k+l-4)(k+l)$ is not positive. The inequality $D \leq 0$ can be easily rearranged to the equivalent condition $k l \geq k+l$.

We have thus found that if the given inequality holds for any triple of side lengths of a triangle, the numbers $k$ and $l$ satisfy, in addition to the inequalities $k \geq 1$ and $l \geq 1$, the condition

$$
k l \geq k+l
$$

From the consideration of the discriminant, it follows conversely that if the numbers $k \geq 1$ and $l \geq 1$ satisfy condition (3), inequality (2) holds for every real $x$. Since for such $k, l$ the value of $V(y)$ is negative for every $y \neq 0$, the validity of (2) for any $x$ implies inequality (1) for all admissible pairs $x, y$, which is equivalent to the given property. The given condition is therefore also sufficient.

The conjunction of the inequalities $k \geq 1, l \geq 1$ and $k l \geq k+l$ is thus a necessary and sufficient condition. Since the third inequality implies $k \neq 1$ (as well as $l \neq 1$), we can write the set of pairs $(k, l)$ that satisfy the condition as follows:

$$
\{(k, l): k>1 \wedge l \geq k /(k-1)\}.
$$