Inequality involving the angle bisectors of a triangle

Question

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.

Answer 1

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.

Answer 2

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 $i=1,2,3$.