Let $a,b,c\ge 0: ab+bc+ca=3$. Prove that$$\frac{1}{\sqrt{5a+4}}+\frac{1}{\sqrt{5b+4}}+\frac{1}{\sqrt{5c+4}}\geq1.$$
This problem is from an AOPS account- arqady. Sorry, I can not find the link.
The equality is $a=b=c=1$, which implies me using $AM-GM:$ $$\sum_{cyc}\frac{1}{\sqrt{5a+4}}\ge 6\sum_{cyc}\frac{1}{5a+13}$$And we need to prove$$\frac{1}{5a+13}+\frac{1}{5b+13}+\frac{1}{5c+13}\ge \frac{1}{6}$$But it is wrong at $a=b=\dfrac{1}{5}$.
I hope to see some ideas. Thank you very much.
By AM-GM, we have $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{5a+4}} = \sum_{\mathrm{cyc}} \frac{2}{2\sqrt{\frac{5a+4}{a + 2}\cdot (a + 2)}} \ge \sum_{\mathrm{cyc}} \frac{2}{\frac{5a + 4}{a + 2} + a + 2}.$$
It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{2}{\frac{5a + 4}{a + 2} + a + 2} \ge 1. \tag{1}$$
We use pqr method.
Let $p = a + b + c, q = ab + bc + ca = 3, r = abc$.
(1) is written as $$-r^2 - (37p + 204)r + 68p + 112 \ge 0. \tag{2}$$
Using $q^2 \ge 3pr$, it suffices to prove that $$- \left(\frac{3}{p}\right)^2 - (37p + 204)\cdot \frac{3}{p} + 68p + 112 \ge 0$$ or $$\frac{(p - 3)(68p + 1)(p + 3)}{p^2} \ge 0$$ which is true using $p^2 \ge 3q$.
We are done.
Another way.
Let $a=\sqrt3\tan\frac{\alpha}{2}$, $b=\sqrt3\tan\frac{\beta}{2}$ and $c=\sqrt3\tan\frac{\gamma}{2}$,where $\{\alpha,\beta,\gamma\}\subset[0,\pi)$.
Thus, $$\sum_{cyc}\tan\frac{\alpha}{2}=1,$$ which gives $\alpha+\beta+\gamma=\pi$ and we need to prove that $\sum\limits_{cyc}f(\alpha)\geq1,$ where $$f(x)=\frac{1}{\sqrt{5\sqrt3\tan\frac{x}{2}+4}}.$$ But $$f''(x)=\frac{\sqrt3(8\sin{x}-10\sqrt3\cos{x}-5\sqrt3)}{\cos\frac{x}{2}\sqrt{\left(5\sqrt3\tan\frac{x}{2}+4\right)^5}},$$ which says that $f$ has on $[0,\pi)$ an unique inflection point $\arcsin\frac{5\sqrt3}{\sqrt{364}}+\arccos\frac{4}{\sqrt{91}}.$
Thus, by the Vasc's HCF Theorem it's enough to prove our inequality for equality case of two variables.
Let $b=a$.
Thus, $c=\frac{3-a^2}{2a},$ where $0<a\leq\sqrt3$ and we need to prove that: $$\frac{2}{\sqrt{5a+4}}+\frac{1}{\sqrt{\frac{5(3-a^2)}{2a}+4}}\geq1$$ or $$4\sqrt{\frac{2(5a+4)(15+8a-5a^2)}{a}}\geq67+30a-25a^2$$ or $$(a-1)^2(384+555a+50a^2-125a^3)\geq0,$$ which is obvious for $0<a\leq\sqrt3.$
About HCF see here: http://ac.upg-ploiesti.ro/membri/vcirtoaje/math_ineq_1.pdf , p6.
Some thoughts.
Remark: The contradiction method helps in another way.
Fact 1. Let $x, y, z \ge 0$ such that $x^2y^2 + y^2z^2 + z^2x^2 + \dfrac{33}{16}x^2y^2z^2 \ge \dfrac{81}{16}$. Then $x + y + z \ge 3$.
Let $$x = \frac{3}{\sqrt{5a + 4}}, \quad y = \frac{3}{\sqrt{5b + 4}}, \quad z = \frac{3}{\sqrt{5c + 4}}.$$
We have $x^2y^2 + y^2z^2 + z^2x^2 + \dfrac{33}{16}x^2y^2z^2 \ge \frac{81}{16}$.
By Fact 1, we have $x + y + z \ge 3$.
Another way.
Let $\sum\limits_{cyc}\frac{1}{\sqrt{5a+4}}<1,$ $a=kp$ such that $k>0$ and $$\frac{1}{\sqrt{5p+4}}+\frac{1}{\sqrt{5b+4}}+\frac{1}{\sqrt{5c+4}}=1.$$ Thus, $$\frac{1}{\sqrt{5kp+4}}+\frac{1}{\sqrt{5b+4}}+\frac{1}{\sqrt{5c+4}}<\frac{1}{\sqrt{5p+4}}+\frac{1}{\sqrt{5b+4}}+\frac{1}{\sqrt{5c+4}},$$ which gives $k>1$ and $$3=ab+ac+bc=k(pb+pc)+bc>pb+pc+bc,$$ which is a contradiction because we'll prove now that $$pb+pc+bc\geq3.$$ Indeed, let $\frac{1}{\sqrt{5p+4}}=\frac{x}{3},$ $\frac{1}{\sqrt{5b+4}}=\frac{y}{3}$ and $\frac{1}{\sqrt{5c+4}}=\frac{z}{3}.$
Thus, $x+y+z=3$, $p=\frac{\frac{9}{x^2}-4}{5},$ $b=\frac{\frac{9}{y^2}-4}{5},$ $c=\frac{\frac{9}{z^2}-4}{5}$ and we need to prove that: $$\sum_{cyc}\left(\frac{\frac{9}{x^2}-4}{5}\cdot\frac{\frac{9}{y^2}-4}{5}\right)\geq3$$ or $$\sum_{cyc}z^2(y+z-x)(x+z-y)(3x+y+z)(3y+x+z)\geq75x^2y^2z^2,$$ where $x$, $y$ and $z$ are sides-lengths of a triangle.
Let $x=v+w$, $y=u+w$ and $z=u+w$, where $u$, $v$ and $w$ are positives.
Thus, we need to prove that: $$16\sum_{cyc}(u+v)^2uv(u+2v+2w)(v+2u+2w)\geq75\prod_{cyc}(u+v)^2$$ or $$\sum_{sym}(32u^5v+69u^4v^2+37u^3v^3+21u^4vw-98u^3v^2w)\geq0,$$ which is true by Muirhead.
