Nesbitt's Inequality

Theorem
Let $a$, $b$ and $c$ be positive real numbers.

Then:
 * $\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$

Proof 1
These are the arithmetic mean and the harmonic mean of $\dfrac 1 {b + c}$, $\dfrac 1 {a + c}$ and $\dfrac 1 {a + b}$.

From Arithmetic Mean is Never Less than Harmonic Mean the last inequality is true.

Thus Nesbitt's Inequality holds.

Proof 2
$$

\begin{align}

\text{Set}\ &S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b},\\

&M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b},\\

&N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}.\\

\text{Obviously,}\ &M+N=1+1+1=3,\\

&M+S=\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{a+c}{a+b}\ge 3,\\

&N+S=\frac{a+c}{b+c}+\frac{a+b}{c+a}+\frac{b+c}{a+b}\ge 3.\\ \\ \end{align} $$ Therefore $M+N+2S\ge 6$, $2S\ge 3$, $S\ge \dfrac{3}{2}$.