<s>[INST] Inequality involving the angle bisectors of a triangle 
Let $l_a,l_b,l_c$ denote the lengths of angle bisectors of a triangle with sides $a,b,c$ and semiperimeter $s$. I am looking for the best constant $K>0$ such that
  $$l_a^2+l_b^2+l_c^2> K s^2.$$

I found that $K=2/3$ works, but I suspect that best constant is $K=8/9>2/3$. Any proof or reference? 
BTW it is known that $l_a^2+l_b^2+l_c^2\leq s^2$.
Proof for $K=2/3$. According to Cut-the-knot, 
$$m_a l_a+m_b l_b+m_c l_c\ge s^{2}$$
where $m_a,m_b,m_c$ are the medians. Therefore, by Cauchy–Schwarz inequality,
$$(m_a^2+m_b^2+m_c^2)(l_a^2+l_b^2+l_c^2)\geq (m_a l_a+m_b l_b+m_c l_c)^2\geq s^4$$
which implies 
$$l_a^2+l_b^2+l_c^2\geq \frac{s^4}{m_a^2+m_b^2+m_c^2}> \frac{2s^2}{3}$$
in view of
$$m_a^{2}+m_b^{2}+m_c^{2}=\frac{3(a^2+b^2+c^2)}{4}< \frac{3s^2}{2}.$$
EDIT. I found a reference that $K=8/9$ is the best constant. See 11.7. at p. 218 in Recent Advances in Geometric Inequalities by Mitrinovic et al.
No proof is given.  
 [/INST] As pointed out by Michael Rozenberg, for $K=8/9$ we have to show that
$$\frac{bc(s-a)}{(b+c)^2}+
\frac{ca(s-b)}{(c+a)^2}+\frac{ab(s-c)}{(a+b)^2}>\frac{2s}{9}.$$
Let $a=x+y$, $b=y+z$, $c=z+x$ with  $x,y,z>0$, then the inequality is equivalent to
$$\sigma_1^3(\sigma_1^2-4\sigma_2)^2+6\sigma_1\sigma_3(\sigma_1\sigma_2
-9\sigma_3)+\sigma_3(19\sigma_1^4+44\sigma_1^2\sigma_2+7\sigma_1\sigma_3+9\sigma_2^2)>0
$$
where $\sigma_1=x+y+z$, $\sigma_2=xy+yz+zx$, $\sigma_3=xyz$.
The above inequality holds because $\sigma_1\sigma_2\geq 9\sigma_3$ by the AM-GM inequality, and $\sigma_i>0$ for $i=1,2,3$.
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