A long time ago, I saw it in a AOPS forum. I've found the original link but it is no longer viewable.
Question. For all $a,b,c>0: a+b+c=ab+bc+ca$. Prove that $$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+33\ge 4\left(\sqrt{\frac{5}{a}+4}+\sqrt{\frac{5}{b}+4}+\sqrt{\frac{5}{c}+4}\right)$$
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 denote $a+b+c=p=ab+bc+ca=q; abc=r.$
By using Cauchy-Schwarz inequality, we'll prove$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+33\ge 4\sqrt{3}\sqrt{\frac{5}{a}+\frac{5}{b}+\frac{5}{c}+12}.$$Or$$\color{black}{33+\frac{q}{r}(q-2r)=33+\frac{q^2-2pr}{r}\ge 4\sqrt{\frac{15q}{r}+36}.}$$
It's easy to see that $p=q\ge 3.$ By squaring both side, it turns out $$513+\left(\frac{q^2-2qr}{r}\right)^2+66\frac{q^2-2qr}{r}-\frac{240q}{r}\ge 0.$$I stopped here since the last inequality is quite complicated.
Also, I tried the old result
$$\color{black}{\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge\sqrt{3(a^2+b^2+c^2)}}.$$
See MSE and more AOPS.
Now, it's enough to prove $$\sqrt{3(a^2+b^2+c^2)}+33\ge 4\left(\sqrt{\frac{5}{a}+4}+\sqrt{\frac{5}{b}+4}+\sqrt{\frac{5}{c}+4}\right).$$
I was stuck here.
Recalled by Michael Rozenberg's proofs, I think the following will help.
By AM-GM, we may find an estimate $$2\sqrt{\frac{5}{a}+4}=2\sqrt{\frac{\dfrac{5}{a}+4}{xa+y}\cdot(xa+y)}\le \frac{\dfrac{5}{a}+4}{xa+y}+(xa+y).$$
We'll choose $x+y=3.$ The rest can be solved by $pqr.$
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
Relevant proof. See my answer in similar answer.
It is also an isolated fudging as RiverLi's answer.
Proof.
By multiplying $ab+bc+ca$ and replace the hypothesis, we'll prove$$\sum_{cyc}\left[\left(\frac{bc}{a}+5\right)(b+c)+8(ab+bc+ca)\right]\ge 4(ab+bc+ca)\sum_{cyc}\sqrt{\frac{5}{a}+4}.$$Now, it's enough to show$$\left(\dfrac{bc}{a}+5\right).\frac{b+c}{ab+bc+ca}+8\ge 4\sqrt{\frac{5}{a}+4}. \tag{*}$$
Apply AM-GM inequality$$\frac{b+c}{ab+bc+ca}=\frac{1}{a+\dfrac{bc}{b+c}}\ge \frac{1}{a+\dfrac{bc}{2\sqrt{bc}}}=\frac{2}{2a+\sqrt{bc}}.$$ It suffices to prove$$\frac{\dfrac{bc}{a}+5}{\sqrt{bc}+2a}+4\ge 2\sqrt{\frac{5}{a}+4}$$ or $$\iff\frac{bc+5a}{\sqrt{bc}+2a}+4a\ge 2\sqrt{5a+4a^2}$$$$\iff \sqrt{bc}+2a+\frac{5a+4a^2}{\sqrt{bc}+2a}\ge 2\sqrt{5a+4a^2},$$which is obvious by AM-GM.
Thus, the proof is done. Equality holds iff $a=b=c=1.$