<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] $\sum\limits_{cyc}\frac{a}{b+c-a}-3=\sum\limits_{cyc}\left(\frac{a}{b+c-a}-1\right)=\sum\limits_{cyc}\frac{a-b-(c-a)}{b+c-a}=$
$=\sum\limits_{cyc}(a-b)\left(\frac{1}{b+c-a}-\frac{1}{a+c-b}\right)=\sum\limits_{cyc}\frac{2(a-b)^2}{(b+c-a)(a+c-b)}\geq0$
</s>