Multiple proofs of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$ Here is my question:

Let $a,b,c\in\mathbb{R^+}$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$$

Here is my solution:
From C-S inequality, we get $$\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq3^2,$$
which is equivalent to $$2(a+b+c) \sum_{cyc}\frac{1}{a+b}\geq9$$
or $$\frac{a+b+c}{a+b}+ \frac{a+b+c}{b+c}+ \frac{a+b+c}{c+a} \geq\frac{9}2$$
or $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{9}2-3=\frac{3}2$$
QED.
Here is my problem:
My teacher said that this question could be done in a lot of ways, one of which is by AM-GM inequality. However, after observing the original inequality, I still couldn’t get a clue.
I am stuck since I completely don’t know where to apply AM-GM. I am thinking about substitute something in the original inequality with some new positive numbers and continue. Is my thought in the right direction?
In addition, other approaches without AM-GM will be appreciated as well. Thanks for help.
P.S.
I am new to Mathematics SE, so if my post can be improved, make sure to let me know through the comment section. Thanks a lot.
 A: One answer using AM-GM.
From AM-GM, we obtain two inequalities:

*

*$a^{\frac{3}2}+b^{\frac{3}2}+b^{\frac{3}2}\geq3a^\frac{1}2b$

*$a^{\frac{3}2}+c^{\frac{3}2}+c^{\frac{3}2}\geq3a^\frac{1}2c$
Add them up and we get:
$$2(a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2})\geq3a^\frac{1}2(b+c)\iff\frac{a}{b+c}\geq\frac{a^\frac{3}2}{a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2}}$$
Therefore
$$\sum_{cyc}\frac{a}{b+c}\geq\frac{3}2\sum_{cyc}\frac{a^\frac{3}2}{a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2}}=\frac{3}2$$QED.
Note: This inequality is called Nesbitt's inequality, and in this link you can see a variety of approaches (as already commented by Dr.Mathva). This approach isn't included in it, so I think you'll probably like this as well.
A: Because
$$\frac{a}{b+c} - \frac{8a-b-c}{4(a+b+c)} = \frac{(2a-b-c)^2}{4(b+c)(a+b+c)} \geqslant 0,$$
so
$$\frac{a}{b+c} \geqslant \frac{8a-b-c}{4(a+b+c)}.$$
Therefore
$$\sum \frac{a}{b+c} \geqslant \sum \frac{8a-b-c}{4(a+b+c)} = \frac 32.$$
Note. In addition, you can see 45th-proof-Nesbitt.pdf (Vietnamese)
A: Using AM-HM, the result can be generalized. Given positive reals $a_1, a_2, \ldots, a_n$, where $n \ge 2$, define
$$b_i = a_1 + \cdots + a_{i-1} + a_{i + 1} + \cdots + a_n$$
and let $A, H$ be the arithmetic and harmonic means of the $b_i$. Then
$$A = (b_1 + \cdots + b_n)/n = (n - 1)(a_1 + \cdots + a_n)/n,$$ whence $a_i = nA/(n-1) - b_i$. Using $A \ge H$ we get
$$\sum \frac{a_i}{b_i} = {\frac{nA}{n-1}}\sum \frac{1}{b_i} - n
= \frac{nA}{n-1}\cdot\frac{n}{H} - n \ge \frac{n^2}{n-1} - n = \frac{n}{n-1}.$$
