Show that the equality $$\dfrac{(b+c)^2}{bc}l_a^2+\dfrac{(c+a)^2}{ca}l_b^2+\dfrac{(a+b)^2}{ab}l_c^2=4p^2$$ holds for a $\triangle ABC$ with sides $AB=c,BC=a, AC=b$, semi-perimeter $p$ and angle bisectors $l_a,l_b$ and $l_c$.
I am a little confused because I have recently seen so many angle bisector formulas. Here we can probably use that $$l_a=\dfrac{2}{b+c}\sqrt{bcp(p-a)}$$ and $$l_b=\dfrac{2}{a+c}\sqrt{acp(p-b)}$$ and $$l_c=\dfrac{2}{a+b}\sqrt{abp(p-c)}.$$ Can you show me? I don't think I can remember these formulas. What can I do if I forget them and can't check because I am not allowed to? What if we don't have them?
The standard angle bisector formula is $$l_a = \frac{2bc \cos\left(\dfrac{A}{2}\right)}{b + c}$$ which, using $\cos\left(\dfrac{A}{2}\right) = \sqrt{\dfrac{p(p - a)}{bc}}$, transforms to $$l_a = \frac{2\sqrt{bcp(p - a)}}{b + c}$$
Now putting these formulas into the LHS, $$ \sum \frac{(b+c)^2}{bc} \left( \frac{2\sqrt{bcp(p - a)}}{b + c} \right)^2 = \sum 4 p(p-a) = 4p \left( \sum (p - a) \right) = 4p^2$$
Hence proved :)