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!
 A: 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
$$\sum_{cyc}(2a^2+ab)\geq3\sum_{cyc}\sqrt[3]{a^2\cdot a^2\cdot bc}=3(a+b+c).$$
A: The friend he told in the problem was me, yeah. And I will post my solution by using the Muirhead Inequality (and also AM-GM Inequality).
$$\dfrac{a}{b+1}+\dfrac{b}{c+1}+\dfrac{c}{a+1}\ge\dfrac{3}{2} \\ \Updownarrow \\ a\left(a+1\right)\left(c+1\right)+b\left(b+1\right)\left(a+1\right)+c\left(c+1\right)\left(b+1\right)\ge\dfrac{3}{2}\left(a+1\right)\left(b+1\right)\left(c+1\right)\\ \Updownarrow \\ a^2c+b^2a+c^2b+a^2+b^2+c^2+ab+bc+ca+a+b+c\ge\\\dfrac{3}{2}abc+\dfrac{3}{2}ab+\dfrac{3}{2}bc+\dfrac{3}{2}ca+\dfrac{3}{2}a+\dfrac{3}{2}b+\dfrac{3}{2}c+\dfrac{3}{2} \\ \Updownarrow (\text{Mutiply by 2 both side and rearranging})\\ 2a^2c+2b^2a+2c^2b+2a^2+2b^2+2c^2\ge3abc+ab+bc+ca+\left(a+b+c\right)a^\frac{1}{3}b^\frac{1}{3}c^\frac{1}{3}+3abc \\ \Updownarrow \\ \left(2a^2c+2b^2a+2c^2b-6abc\right)+\sum_{sym}\left(a^2-\dfrac{1}{2}ab-\dfrac{1}{2}a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}\right)\ge0 $$ By AM-GM Inequality, $$2a^2c+2b^2a+2c^2b\ge 3\sqrt[3]{8a^3b^3c^3}=6abc \\ \therefore 2a^2c+2b^2a+2c^2b-6abc\ge 0$$
By Muirhead Inequality,
$$\because \left(2,0,0\right)\succ\left(1,1,0\right),\left(2,0,0\right)\succ\left(\dfrac{4}{3},\dfrac{1}{3},\dfrac{1}{3}\right) \\ \therefore \sum_{sym}\left(a^2-\dfrac{1}{2}ab-\dfrac{1}{2}a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}\right)=\dfrac{1}{2}\sum_{sym}\left(a^2-ab\right)+\dfrac{1}{2}\sum_{sym}\left(a^2-a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}\right)\ge0$$
Adding the two inequality we prove, we get the answer.
