How prove $\dfrac{|AB|}{|AC|}\le\sqrt{2}+1$

Question

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 $$\dfrac{AB}{AC}=\dfrac{\sin{C}}{\sin{B}}$$

Answer 1

I think you have only to check the worst of the cases, so is, prove that $\sqrt{5}\le\sqrt 2 +1$

Answer 2

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.

Answer 3

First of all, I must point out that the picture is wrongly drawn. The reason is if AB is shorter than AC, the LHS of the inequality-to-be-proved is less than 1 and obviously is less than $\sqrt 2 + 1$, the RHS. This means there is nothing to be proved. Therefore, we must have $AB \ge AC$ ….. (1).

If we let $AD = 1$ and $DC = x$ for some x greater 0, then $BD = 2 – x$ and $AC = \sqrt {1 + x^2}$.

(1) implies $\sqrt {(2 – x)^2 + 1} \ge \sqrt {1 + x^2}$.

After simplifying, we arrive at the first conclusion --- $0 \le x \le 1$. Then, $AC_{min} = 1$ ….. (2).

After some simplification. the inequality-to-be-proved is equivalent to $\sqrt 2 AC \ge BC’$.

It will be true if we can prove $\sqrt 2AC \ge \sqrt 2 AC_{min} \ge BC’_{max} \ge BC’$.

Or simply, $\sqrt 2 AC_{min} \ge BC’_{max}$ …. (*)

From (2), LHS = $\sqrt 2$ …. (3)

RHS = Max $[\sqrt {(2 – x)^2 + 1} - \sqrt {1 + x^2}] = Max[\sqrt {(2 – x)^2 + 1}] – Min [\sqrt {1 + x^2}] = ....$

$ =\sqrt 5 – 1$ …. (4)

(*) is true by comparing (3) and (4).