in right triangle

Let $|\overline {AB}|=c, |\overline {BC}|=a$ . Then $|\overline {AC}|=\sqrt {a^2+c^2}$ . Since $a, c$ are positive reals, we apply the R. M. S. - A. M. inequality to get $\sqrt {\dfrac{a^2+c^2}{2}}\geq \dfrac{a+c}{2}\implies \dfrac{a+c}{\sqrt {a^2+c^2}}\leq \sqrt 2$ . Hence $m=\sqrt 2\implies m^2=\boxed 2$ . There arises a confusion with the symbols the author of this problem used. By $ab$ he didn't mean $a\times b$ , but $|\overline {AB}|$ . Similar for others.

Let $|CA|=1$ and $\angle A = \theta$ . Then $|AB|=\cos \theta$ and $|BC|=\sin \theta$ and

$\begin{aligned} \frac {|AB|+|BC|}{|CA|} & = \frac {\cos \theta + \sin \theta}1 = \sqrt 2 \blue{\sin \left(\theta + \frac \pi4\right)} \le \sqrt 2 & \small \blue{\text{Since }-1 \le \sin \phi \le 1} \end{aligned}$

Therefore, $m^2 = \boxed 2$ .