A geometric inequality for a triangle ABC

Question

I have to prove that: $ \frac {a^2}{w_a^2} + \frac {b^2}{ w_b^2} +\frac {c^2}{ w_c^2} \ge 4,$

for the sides $a,b,c$ of the triangle $ABC,$ $ w_a, w_b, w_c $ the angle bisectors and $s$ its semiperemeter. We have for the angle bisectors of the triangle $ABC$:

$ w_a = \frac {2}{b+c} \sqrt{b c s (s-a)},$ $ w_b = \frac {2}{c+a} \sqrt{c a s (s-b)},$ $ w_c = \frac {2}{a+b} \sqrt{a b s (s-c)}.$

Then, we have:

$ \frac {a^2}{w_a^2} + \frac {b^2}{ w_b^2} +\frac {c^2}{ w_c^2} = \frac {1}{4 s} [ \frac { a^2 (b+c)^2}{b c (s-a)} + \frac { b^2 (a+c)^2}{a c (s-b)} + \frac { c^2 (a+b)^2}{a b (s-c)}].$

So, we have to prove

$ \frac{ a^2 (b+c)^2}{b c (s-a)} + \frac { b^2 (a+c)^2}{a c (s-b)} + \frac { c^2 (a+b)^2}{a b (s-c)} \ge 16 s $

Then I stuck, thank you.

Answer 1

Taking from where you left off... $LHS \ge \sum 4\cdot\dfrac{a^2}{s-a}\ge 4\cdot\dfrac{(a+b+c)^2}{(s-a)+(s-b)+(s-c)}= 4\cdot\dfrac{4s^2}{s} = 16s = RHS$ , by employing AM-GM and CS inequalities.

Answer 2

Let $m_a$, $m_b$ and $m_c$ be medians of the triangle to sides $a$, $b$ and $c$ respectively.

Thus, by C-S and AM-GM we obtain: $$\sum_{cyc}\frac{a^2}{w_a^2}\geq\sum_{cyc}\frac{a^2}{m_a^2}=4\sum_{cyc}\frac{a^2}{2b^2+2c^2-a^2}=4\sum_{cyc}\frac{a^4}{2a^2b^2+2a^2c^2-a^4}\geq$$ $$\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+2a^2c^2-a^4)}\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^4+c^4-a^4)}=\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^4)}=4.$$