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!
 A: Let $a=y+z$, $b=x+a$ and $c=x+y$.
Thus,$x$, $y$ and $z$ are positives and we need to prove that:
$$4(x+y+z)\sum_{cyc}\frac{1}{x+y}-9\geq\sum_{cyc}\sqrt{17-\frac{16yz(2x+y+z)}{\prod\limits_{cyc}(x+y)}}$$ or
$$\sum_{cyc}(4x^3+7x^2y+7x^2z+6xyz)\geq\sum_{cyc}\sqrt{\prod_{cyc}(x+y)(17\prod_{cyc}(x+y)-16yz(2x+y+z))}.$$
Now, by AM-GM
$$\sum_{cyc}\sqrt{\prod_{cyc}(x+y)(17\prod_{cyc}(x+y)-16yz(2x+y+z))}=$$
$$=\frac{1}{3}\sum_{cyc}\sqrt{9\prod_{cyc}(x+y)\cdot(17\prod_{cyc}(x+y)-16yz(2x+y+z))}\leq$$
$$\leq\frac{1}{6}\sum_{cyc}\left(9\prod_{cyc}(x+y)+17\prod_{cyc}(x+y)-16yz(2x+y+z)\right)$$ and it's enough to prove that:
$$6\sum_{cyc}(4x^3+7x^2y+7x^2z+6xyz)\geq\sum_{cyc}\left(9\prod_{cyc}(x+y)+17\prod_{cyc}(x+y)-16yz(2x+y+z)\right)$$ or
$$\sum_{cyc}(6x^3-5x^2y-5x^2z+4xyz)\geq0$$ or
$$5\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\sum_{cyc}(x^3-xyz)\geq0,$$ which is true by Schur and Muirhead.
A: 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: $\frac{c}{2(a+b)-c}+\frac{a}{2(b+c)-a}+\frac{b}{2(c+a)-b}\ge1$
