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.
 A: $a, b, c$ are sides of a triangle iff there exists positive reals $x, y, z$ s.t. $a=x+y, b=y+z, c = z+x$.  In terms of these variables, the inequality is
$$\sum_{cyc} \frac{a}{b+c-a} = \sum_{cyc} \frac{x+y}{2z} \ge 3$$
Now the last is easy to show with AM-GM of all $6$ terms.
$$\sum_{cyc} \frac{x+y}{2z} = \frac12\left(\frac{x}z+\frac{y}z+\frac{y}x+\frac{z}x+\frac{z}y+\frac{x}y \right) \ge \frac12\left(6\sqrt[6]{\frac{x}z\cdot\frac{y}z\cdot\frac{y}x\cdot\frac{z}x\cdot\frac{z}y\cdot\frac{x}y} \right) = 3$$
A: 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 $\ $ $\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\geq 3$
A: By C-S $\sum\limits_{cyc}\frac{a}{b+c-a}=\sum\limits_{cyc}\frac{a^2}{ab+ac-a^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2ab-a^2)}\geq3$, where the last inequality it's just
$$\sum\limits_{cyc}(a-b)^2\geq0.$$
Done!
A: $\sum\limits_{cyc}\frac{a}{b+c-a}-3=\sum\limits_{cyc}\left(\frac{a}{b+c-a}-1\right)=\sum\limits_{cyc}\frac{a-b-(c-a)}{b+c-a}=$
$=\sum\limits_{cyc}(a-b)\left(\frac{1}{b+c-a}-\frac{1}{a+c-b}\right)=\sum\limits_{cyc}\frac{2(a-b)^2}{(b+c-a)(a+c-b)}\geq0$
A: Since $(a,b,c)$ and $\left(\frac{1}{b+c-a},\frac{1}{a+c-b},\frac{1}{a+b-c}\right)$ are the same ordered, by Chebyshov and C-S we obtain:
$\sum\limits_{cyc}\frac{a}{b+c-a}\geq\frac{1}{3}(a+b+c)\sum\limits_{cyc}\frac{1}{b+c-a}=\frac{1}{3}\sum\limits_{cyc}(b+c-a)\sum\limits_{cyc}\frac{1}{b+c-a}\geq\frac{1}{3}\cdot9=3$.
A: Suppose that $S>0$. Then for $x\in(0,S/2)$, the function $f(x)=\frac{x}{S-2x}$ is convex. Thus, by Jensen's inequality and with $S=a+b+c$, we have
$$
\frac{1}{3}f(a)+\frac{1}{3}f(b)+\frac{1}{3}f(c)\geq f\left(\frac{1}{3}(a+b+c)\right)=\frac{S/3}{S-2S/3}=1.
$$
This is equivalent to $f(a)+f(b)+f(c)\geq 3$, which is your inequality.
A: 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}
