Find a nice proof for an elegant problem.

Question

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.

Answer 1

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 $a=b=c>0.$

Answer 2

We need to prove that: $$\sum_{cyc}a\sqrt{5a^2+4bc}\leq\frac{9}{4}\left(a^2+b^2+c^2+\frac{abc(a+b+c)}{ab+ac+bc}\right).$$ Now, by C-S $$\sum_{cyc}a\sqrt{5a^2+4bc}=\sqrt{\sum_{cyc}\left(5a^4+4a^2bc+2ab\sqrt{(5a^2+4bc)(5b^2+4ac)}\right)}\leq$$ $$\leq\sqrt{\sum_{cyc}(5a^4+4a^2bc)+2\sqrt{(ab+ac+bc)\sum_{cyc}ab(5a^2+4bc)(5b^2+4ac)}}$$ and it's enough to prove that: $$\sum_{cyc}(5a^4+4a^2bc)+2\sqrt{(ab+ac+bc)\sum_{cyc}ab(5a^2+4bc)(5b^2+4ac)}\leq$$ $$\leq\frac{81}{16}\left(a^2+b^2+c^2+\frac{abc(a+b+c)}{ab+ac+bc}\right)^2,$$ which is true, but my proof of this statement is not so nice yet.