**Problem:** Given a,b,c are length of triangle. Prove that: $$2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum\_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$
*Happy Vietnamese Women's Day*

Phan Ngoc Chau, Oct 20th 2021

*My approach:* Note that: $(b-c)^2-a^2=(b-c-a)(b-c+a)<0$ so $\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}<\sqrt{17}$

And as well- known result: $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9$

But it is not true to show: $3\sqrt{17}\le9$. It is quite ugly for me to get a proof. Anyone help me ? Thanks!