Olympiad Inequality with Condition

Question

I would like to prove this :

Let $x,y,z$ be positive real numbers such that $xyz=1$ then we have : $$\frac{\left(\frac{1}{x}+\frac{z}{x}+z\right)\left(1+\frac{1}{x}+z\right)}{3\left(\frac{z}{x}\right)}+\sum_{cyc}^{}\sin\left(\frac{\sqrt{\left(1+\frac{1}{x}+z\right)}}{\sqrt{3\left(\frac{z}{x}\right)}}-1\right)\geq 3$$

My try :

Remark that $xyz=1$ is equivalent to $$\frac{1}{1+\frac{1}{x}+z}+\frac{1}{1+\frac{1}{y}+x}+\frac{1}{1+\frac{1}{z}+y}=1$$

And :

$$\frac{\left(\frac{1}{x}+\frac{z}{x}+z\right)\left(1+\frac{1}{x}+z\right)}{3\left(\frac{z}{x}\right)}=\frac{\left(\frac{1}{y}+\frac{x}{y}+x\right)\left(1+\frac{1}{y}+x\right)}{3\left(\frac{x}{y}\right)}=\frac{\left(\frac{1}{z}+\frac{y}{z}+y\right)\left(1+\frac{1}{z}+y\right)}{3\left(\frac{y}{z}\right)}$$

But after this I have no idea to prove this ...

Thanks in Advance .

Answer 1

Maybe the following will help.

Let $x=\frac{a}{b}$, $y=\frac{b}{c}$ and $z=\frac{c}{a}$, where $a$, $b$ and $c$ are positives.

Thus, we need to prove that $$\frac{(a+b+c)(ab+ca+bc)}{3abc}+\sum_{cyc}\sin\left(\sqrt{\frac{(a+b+c)a}{3bc}}-1\right)\geq3.$$