How prove $\dfrac{|AB|}{|AC|}\le\sqrt{2}+1$ in $\Delta ABC$ if the $AD\perp BC$,$D\in BC$,and such $$|BC|=2|AD|$$
show that
$$\dfrac{|AB|}{|AC|}\le\sqrt{2}+1$$


since
$$\cot{B}+\cot{C}=\dfrac{BD}{AD}+\dfrac{CD}{AD}=2$$
so
$$\dfrac{AB}{AC}=\dfrac{\sin{C}}{\sin{B}}$$
This is more of an algebra problem in fact, and I misled myself into thinking of Stewart's theorem at first.  In fact if we let the length of $AD$ be 1 and $BC$ be 2 it becomes equivalent to trying to maximise an expression like $\frac{x^2+1}{(2-x)^2+1}$ where $x$ varies between $0$ and $2$  and since the numerator increases while the denominator decreases as $x$ increases it suffices to let $x=2$ giving a bound of $\sqrt{5}$ which is smaller than the given bound.  I do not know if there is another method that gives $\sqrt{2}+1$ though.
I think you have only to check the worst of the cases, so is, prove that $\sqrt{5}\le\sqrt 2 +1$