Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$. 
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$

A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.

Since $a,b,c$ are sides of a triangle, we can set $a = x+y$, $b = x+z$, $c = y+z$.
Plugging that in gives
\begin{align}
\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} &= \frac{x+y}{2z} + \frac{x+z}{2y} + \frac{y+z}{2x} \\
&= \frac{xy(x+y) + xz(x+z) + yz(y+z)}{2xyz}\\
&= \frac{x^2y+ xy^2 + x^2z + xz^2 + y^2z + yz^2}{2xyz}\\
&\ge \frac{3}{xyz}\sqrt[6]{x^6y^6z^6}\\
&= 3
\end{align}
