Saudi Arabia booklet, 2024

The given inequality can be written as
$$2k \cdot \frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{(a + b + c)^2} \leq \frac{\sqrt{a^2 + b^2 + c^2} - \sqrt{ab + bc + ca}}{\sqrt{a^2 + b^2 + c^2}}$$
$$2k \cdot \frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{(a + b + c)^2} \leq \frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{\sqrt{a^2 + b^2 + c^2}((a^2 + b^2 + c^2) + (ab + bc + ca))}.$$
Note that
$$(a^2 + b^2 + c^2) - (ab + bc + ca) = \frac{1}{2} \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right] \geq 0$$
so
$$2k\sqrt{a^2 + b^2 + c^2} \left( \sqrt{a^2 + b^2 + c^2} + \sqrt{ab + bc + ca} \right) \leq (a + b + c)^2.$$
Consider $a = b = 1$ and $c \to 0^+$ (isosceles triangle with the base arbitrary small) then $\text{LHS} \to 2k \cdot \sqrt{2}(\sqrt{2} + 1)$ and $\text{RHS} \to 4$. Thus
$$k \leq \frac{4}{2\sqrt{2}(\sqrt{2} + 1)} = 2 - \sqrt{2}.$$
For $k = 2 - \sqrt{2}$ then we need to prove that
$$(4 - 2\sqrt{2})\sqrt{a^2 + b^2 + c^2} \left( \sqrt{a^2 + b^2 + c^2} + \sqrt{ab + bc + ca} \right) \leq (a + b + c)^2.$$
Denote $x = \sqrt{a^2 + b^2 + c^2}$ and $y = \sqrt{ab + bc + ca}$ then the above can be written as
$$(4 - 2\sqrt{2})x(x + y) \leq x^2 + 2y^2$$
$$x^2 + 2(2 + \sqrt{2})xy - (2 + \sqrt{2})^2 y^2 \leq 0.$$
This equivalent to $x \leq y\sqrt{2}$ or $a^2 + b^2 + c^2 \leq 2(ab + bc + ca)$. The last inequality is true for $a, b, c$ are sidelengths of triangle since it can be written as
$$a(b + c - a) + b(c + a - b) + c(a + b - c) \geq 0.$$
Hence, $k_{\max} = 2 - \sqrt{2}$.