<s>[INST] 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.
 [/INST] Set $b+c-a=x, a+c-b=y$ and $a+b-c=z$
Due to triangular inequality, $x,y,z$ are positive. Hence we can apply the AM-GM inequality.
We have $x+y=b+c-a+a+c-b=2c \implies c = \frac{x+y}{2}$
Similarly $a = \frac{y+z}{2}$ and $b = \frac{z+x}{2}$
Now we have,
$$\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}=\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z}$$
$$\Leftrightarrow 2\left ( \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} \right )=\left ( \frac{x}{y}+\frac{y}{x} \right )+\left ( \frac{y}{z}+\frac{z}{y} \right )+\left ( \frac{z}{x}+\frac{x}{z} \right )\geq 6$$
Hence $\ $ $\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\geq 3$
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