Irska

Let $D=D(\lambda )$ for a fixed $\lambda$ and denote $\theta =\angle ADB=180^{\circ }-\angle CDA$. The Cosine Rule for the triangles $△DAB$ and $△CAD$ gives $$\begin{array}{rl}{c}^2& =|BD{|}^2+|AD{|}^2-2|BD|\cdot |AD|\cos \theta \\ b^2& =|DC{|}^2+|AD{|}^2+2|DC|\cdot |AD|\cos \theta .\end{array}$$ 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 $$|BD|=c^{\lambda }\frac{a}{b^{\lambda }+c^{\lambda }}\text{and}|DC|=b^{\lambda }\frac{a}{b^{\lambda }+c^{\lambda }}.$$ 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 $\lambda \ge 1$.