Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$?

I'm sorry but AM-GM Inequality is all that I know, So, By using AM-GM Inequality for $ (a+b) , (b+c) , (c+a)$ and $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ We get $$\frac{(a+b)+(b+c)+(c+a)}{3} \ge\sqrt[3]{(a+b)(b+c)(c+a)}\tag{1}$$ $$\frac{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}{3} \ge\sqrt[3]{\frac{1}{(b+c)(c+a)(a+b)}}\tag{2}$$ Multiplying $(1)$ and $(2)$ We get. $$\frac{1}{9}\left[(a+b)+(b+c)+(c+a)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{(b+c)(c+a)(a+b)}}$$ $$\left[(a+b)+(b+c)+(c+a)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9$$ After Expanding $$3+\frac{a+b}{b+c}+\frac{c+a}{b+c}+ \frac{a+b}{c+a}+\frac{b+c}{c+a}+ \frac{b+c}{a+b}+\frac{c+a}{a+b} \ge\ 9$$ By rearrangement $$\left(\frac{a+b}{b+c}+\frac{c+a}{b+c}\right)+ \left(\frac{a+b}{c+a}+\frac{b+c}{c+a}\right)+ \left(\frac{b+c}{a+b}+\frac{c+a}{a+b}\right) \ge\ 6$$ $$\frac{2a+b+c}{b+c}+ \frac{a+2b+c}{c+a}+ \frac{b+2c+a}{a+b} \ge\ 6$$ $$\frac{2a}{b+c}+\frac{b+c}{b+c}+ \frac{2b}{c+a}+\frac{a+c}{c+a}+ \frac{2c}{a+b}+\frac{b+a}{a+b} \ge\ 6$$ $$3+\frac{2a}{b+c}+ \frac{2b}{c+a}+ \frac{2c}{a+b} \ge\ 6$$ $$\frac{2a}{b+c}+ \frac{2b}{c+a}+ \frac{2c}{a+b} \ge\ 3$$ $$\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b} \ge\ \frac{3}{2}$$ As Simple as that

First Proof

$a+b=x$ , $b+c=y$ , $c+a=z$

$\therefore 2 \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=(x+y+z)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-6\\=\underbrace{\dfrac{x}{y}+\dfrac{y}{x}}_{\geq 2}+\underbrace{\dfrac{y}{z}+\dfrac{z}{y}}_{\geq 2}+\underbrace{\dfrac{z}{x}+\dfrac{x}{z}}_{\geq 2}-3 \geq 3$

Second Proof

$f(a,b,c)=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right) \implies f(ta,tb,tc)=f(a,b,c)$ $\forall t\in \mathbb{R}\setminus\{0\}$ $a+b+c=1 \implies f(a,b,c)=\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3 \geq \dfrac{9}{2}-3=\dfrac{3}{2}$

If you want to use the power series way to do it(although I think it's overkill), by convex property of $f(x)=x^n$ we have $\frac{(a^n+b^n+c^n)}{3}\geq (\frac{a+b+c}{3})^n$ or equivalently $(a^n+b^n+c^n)\geq \frac{(a+b+c)^n}{3^{n-1}}$, where $n$ is positive integer.

Therefore your left hand side(series expansion) is greater or equal to $1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}...=\frac{3}{2}$