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.  

Let $a=b=1$ and $c\rightarrow2^-$.
Thus, $K<\frac{8}{9}.$
We'll prove that $\frac{8}{9}$ it's a best constant. 
Indeed, we need to prove that:
$$\sum_{cyc}\left(\frac{2bc\cos\frac{\alpha}{2}}{b+c}\right)^2\geq\frac{8}{9}\cdot\frac{(a+b+c)^2}{4}$$ or
$$\sum_{cyc}\left(\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}{b+c}\right)^2\geq\frac{2(a+b+c)^2}{9}$$ or
$$\sum_{cyc}\frac{bc(b+c-a)}{(b+c)^2}\geq\frac{2(a+b+c)}{9}.$$
Now, let $a=x+u$, $b=x+v$ and $c=x+u+v,$ where $x>0$ and $u$ and $v$ are non-negatives.
Thus, we need to prove that:
$$48x^7+224(u+v)x^6+16(23u^2+61uv+23v^2)x^5+$$
$$+16(u+v)(16u^2+75uv+16v^2)x^4+$$
$$+(65u^4+894u^3v+1859u^2v^2+894uv^3+65v^4)x^3+$$
$$+(4u^5+185u^4v+900u^3v^2+900u^2v^3+185uv^4+4v^5)x^2+$$
$$+(5u^6-7u^5v+121u^4v^2+275u^3v^3+121u^2v^4-7uv^5+5v^6)x+$$
$$+2(u+v)^3(u-v)^4\geq0,$$
which is obvious.
