I made an inequality to solve myself, but someone pointed out my solution is wrong. I made the following inequality for myself to solve, but my friend found out an mistake:

If $a,b,c\in\mathbb{R^{+}}$ and $abc=1$, prove that $$\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}\ge\frac32$$

I tried to substitute $a=\frac xy, b= \frac yz, c= \frac zx$, but that friend found out a mistake. He wrote a solution using Muirhead’s inequality, but that is not what I want. I also tried trigonometric substitution, but it turned out that there is a mistake also. Is there any method which doesn’t use Muirhead’s inequality? Thanks for any help!

By C-S $$\sum_{cyc}\frac{a}{b+1}=\sum_{cyc}\frac{a^2}{ab+a}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+a)}.$$
Thus, it's enough to prove that
$$2(a+b+c)^2\geq3\sum_{cyc}(ab+a)$$ or
$$\sum_{cyc}(2a^2+ab)\geq3(a+b+c).$$
Now, by AM-GM $$\sum_{cyc}ab\geq3\sqrt[3]{a^2b^2c^2}=3$$ and by C-S again
$$a^2+b^2+c^2=\frac{1}{3}(1+1+1)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2.$$
Thus, it's enough to prove that
$$\frac{2}{3}(a+b+c)^2+3\geq3(a+b+c)$$ or
$$(2(a+b+c)-3)(a+b+c-3)\geq0.$$
Can you end it now?
Also, by AM-GM
$$\sum_{cyc}(2a^2+ab)\geq3\sum_{cyc}\sqrt[3]{a^2\cdot a^2\cdot bc}=3(a+b+c).$$
