In triangle. Prove that: $2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$ 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!

I found a stronger one:
Problem: Given a,b,c are length of triangle. Prove that: $$2\sum_{cyc}{\frac{a+b-c}{c}}\sum_{cyc}{\frac{a+b-c}{2(a+b)-c}}+3\ge\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$
Note that: $\frac{c}{2(a+b)-c}+\frac{a}{2(b+c)-a}+\frac{b}{2(c+a)-b}\ge1$
