Problem 1261

Let $D = D(\lambda)$ for a fixed $\lambda$ and denote $\theta = \angle ADB = 180^\circ - \angle CDA$. The Cosine Rule for the triangles $\triangle DAB$ and $\triangle CAD$ gives
$$\begin{aligned} c^2 &= |BD|^2 + |AD|^2 - 2|BD| \cdot |AD| \cos \theta \\ b^2 &= |DC|^2 + |AD|^2 + 2|DC| \cdot |AD| \cos \theta. \end{aligned}$$

Eliminating $\cos \theta$ gives
$$\frac{|BD|^2 + |AD|^2 - c^2}{2|BD| \cdot |AD|} = \frac{b^2 - |DC|^2 - |AD|^2}{2|DC| \cdot |AD|}$$
from which we obtain
$$|AD|^2 = \frac{b^2|BD| + c^2|DC|}{|BD| + |DC|} - |BD| \cdot |DC|.$$
From $|BD|/|DC| = c^\lambda / b^\lambda$ and $|BD| + |DC| = a$ we deduce
$$|BD| = c^{\lambda} \frac{a}{b^{\lambda} + c^{\lambda}} \quad \text{and} \quad |DC| = b^{\lambda} \frac{a}{b^{\lambda} + c^{\lambda}}.$$
Substitution into the previous formula establishes that
$$|AD(\lambda)|^2 = \frac{b^{\lambda}c^{\lambda}}{(b^{\lambda} + c^{\lambda})^2} ((b^{2-\lambda} + c^{2-\lambda})(b^{\lambda} + c^{\lambda}) - a^2).$$
Hence
$$\frac{|AD(\lambda)|^2}{|AD(2-\lambda)|^2} = \frac{b^{\lambda}c^{\lambda}}{(b^{\lambda} + c^{\lambda})^2} \frac{(b^{2-\lambda} + c^{2-\lambda})^2}{b^{2-\lambda}c^{2-\lambda}},$$
so that
$$\frac{|AD(\lambda)|}{|AD(2-\lambda)|} = \frac{b^{\lambda-1}c^{\lambda-1}(b^{2-\lambda} + c^{2-\lambda})}{b^{\lambda} + c^{\lambda}} = \frac{t^{\lambda-1}(t^{2-\lambda} + 1)}{t^{\lambda} + 1},$$
where $t = b/c$. Hence $|AD(\lambda)| \le |AD(2 - \lambda)|$ iff
$$t + t^{\lambda-1} \le t^{\lambda} + 1$$
equivalently iff
$$0 \le 1 - t + t^{\lambda-1}(t-1) = (1-t)(1-t^{\lambda-1})$$
which holds for $t > 0$ iff $\lambda \ge 1$.