Question. For any $a,b,c>0$ then prove that$$\frac{\sqrt{5a^2+4bc}}{bc}+\frac{\sqrt{5b^2+4ca}}{ca}+\frac{\sqrt{5c^2+4ab}}{ab}\le \frac{9}{4}\left(\frac{a^2+b^2+c^2}{abc}+\frac{a+b+c}{ab+bc+ca}\right).$$
Source: unknown.
Context.
A long time ago, I saw it in a group Facebook. I've found the original link but it is no longer viewable.
I want to find some elegant proofs for this nice problem. It would be great if there's an AM-GM or Cauchy-Schwarz ideas.
Firstly, I verified it by multiply $abc$ and the problem becomes $$4\left(a\sqrt{5a^2+4bc}+b\sqrt{5b^2+4ac}+c\sqrt{5c^2+4ab}\right)\le 9\left(a^2+b^2+c^2+\frac{abc(a+b+c)}{ab+bc+ca}\right). $$
Naturally, we can eliminate the square root of left side by C-S.
It's $$a\sqrt{5a^2+4bc}+b\sqrt{5b^2+4ac}+c\sqrt{5c^2+4ab}\le \sqrt{(a^2+b^2+c^2)(5a^2+5b^2+5c^2+4ab+4bc+4ca)} \tag{1}$$$$a\sqrt{5a^2+4bc}+b\sqrt{5b^2+4ac}+c\sqrt{5c^2+4ab}\le \sqrt{(a+b+c)(5a^3+5b^3+5c^3+12abc)} \tag{2}$$
I checked $a=b=0.99; c=1$ so $(1)$ and $(2)$ lead to wrong inequalities.
Also, we can come up with (in some ways) an isolated fuding inspired by RiverLi's idea for similar inequality.
Indeed, we may find an estimate$$\color{black}{\frac{xb+xc+ya}{ab+bc+ca}+\frac{9a}{bc}\ge 4\frac{\sqrt{5a^2+4bc}}{bc} .}$$
In cases there's exist $(x,y),$ I hope the solver will give your motivation to obtain that approach.
That's all what I've done so far. Thank you for your contributions.
Best Answer
Remark.
The following proof is inspired by Michael Rozenberg's idea. See topic.
The motivation is based on $\sum_{sym}\dfrac{b+c}{2(ab+bc+ca)}\ge \sum_{sym} \dfrac{1}{2a+\sqrt{bc}}.$
Proof.
We may use AM-GM for $2a+\sqrt{bc}$ as $$2\sqrt{5a^2+4bc}=2\sqrt{\frac{5a^2+4bc}{2a+\sqrt{bc}}\cdot(2a+\sqrt{bc})}\le \frac{5a^2+4bc}{2a+\sqrt{bc}}+2a+\sqrt{bc}=4a+\frac{5bc+a^2}{2a+\sqrt{bc}}.$$ Also, $$2a+\sqrt{bc}\ge 2a+\dfrac{2bc}{b+c}=\dfrac{2(ab+bc+ca)}{b+c}.$$ Thus, we obtain $$4\frac{\sqrt{5a^2+4bc}}{bc}\le \frac{8a}{bc}+\frac{(b+c)\left(\dfrac{a^2}{bc}+5\right)}{ab+bc+ca}.$$Notice that $$(b+c)\left(\dfrac{a^2}{bc}+5\right)=(ab+ac+bc)\frac{a}{bc}+5(b+c)-a.$$ And we see $$\frac{5(b+c)-a}{ab+bc+ca}+\frac{9a}{bc}\ge 4\frac{\sqrt{5a^2+4bc}}{bc} . $$ Take cyclic sum on it, we get desired inequality. Equality holds iff $a=b=c>0.$