Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.
"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.

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
