I can't say this is a good problem, but I think the manipulation behind it is fairly mysterious so if anyone is interested, here are some hints:

After some serious prodding from math154, here is my attempt at filling in the details: Consider three mutually (externally) tangent circles with centers $X, Y, Z$ of radii $1/x, 1/y, 1/z$ respectively; the triangle formed from their centers has sides $1/x + 1/y$, $1/x + 1/z$, $1/y + 1/z$ and so its incircle has curvature \[\frac{1}{r} = \sqrt{\frac{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}{\frac{1}{x} \cdot \frac{1}{y} \cdot \frac{1}{z}}} = \sqrt{xy + yz + zx} = 1\] Denote by $s$ the curvature of the circle externally tangent to these three circles, and by $S$ its center. By Descartes' Theorem, \[s = x + y + z + 2\sqrt{xy + yz + zx}\] Let $t$ be the unique positive real satisfying \[ \sum_{\text{cyc}}\sqrt{(xy+tx+ty)(xz+kx+kz)} = t^2\] It's sufficient to show $k \le t$. Set $k_x = \sqrt{yz + ty + tz}$ and analogously $k_y, k_z$, so that $k_x$ is the curvature of the incircle of $\triangle{SYZ}$ and so on. By equal tangents, the incircles of triangles $SYZ$, $SXZ$, and $SXY$ are mutually externally tangent and all internally tangent to the incircle of $\triangle{XYZ}$; by Descartes' Theorem again, \[k_x + k_y + k_z - 2\sqrt{k_xk_y + k_yk_z + k_zk_x} = -1\] Moving the radical, squaring, and canceling \[k_x^2 + k_y^2 + k_z^2 = 2k_xk_y + 2k_yk_z + 2k_zk_x - 4\sqrt{k_xk_y + k_yk_z + k_zk_x} + 1\] \[(xy + yz + xz) + 2t(x + y + z) = 2(k_xk_y + k_yk_z + k_zk_x) - 4\sqrt{k_xk_y + k_yk_z + k_zk_x} + 1\] \[2t(x + y + z) = 2t^2 - 4t\] \[t = x + y + z + 2 = x + y + z + 2\sqrt{xy + yz + zx} = s\] Now $xy + yz + zx = 1$ so $x + y + z \ge \sqrt{3}$; it follows that \[t = x + y + z + 2\sqrt{xy + yz + zx} = x + y + z + 2 \ge 2 + \sqrt{3} = k\]