Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$ 
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equilateral.

$\bf{My\; Try::}$ We can write it as $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1$$
So $$\left(\frac{a+b-c}{a}\right)^{a}\cdot \left(\frac{b+c-a}{b}\right)^{b}\cdot \left(\frac{c+a-b}{c}\right)^{c}\geq 1$$
So $$(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$$
Now how can i solve after that, Help required, Thanks
 A: The title's inequality is wrong. I'll prove a reversed inequality.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Since our inequality does not changed after substitution $a\rightarrow ka$...,
we can assume that $x+y+z=3$.
And we need to prove that $\sum\limits_{cyc}(3-x)\ln\frac{2x}{3-x}\leq0$ or
$\sum\limits_{cyc}\left(3(x-1)-(3-x)\ln\frac{2x}{3-x}\right)\geq0$, which is true because
$3(x-1)-(3-x)\ln\frac{2x}{3-x}\geq0$ is true for all $x\in(0,3)$.
The equality occurs for $x=y=z=1$.
Thus, $a=b=c$ and we are done!
Maybe you mean the following problem.

If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equilateral.

Let $a+b+c=1$.
Hence, by Jensen $a\ln\left(1+\frac{b-c}{a}\right)\leq\ln\sum\limits_{cyc}(a+b-c)=1$.
Thus, $\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\leq 1$ and with the given we obtain $a=b=c$.
A: I will prove $a^a\cdot b^b\cdot c^c \geq (a+b-c)^{c}\cdot (b+c-a)^{b}\cdot (c+a-b)^{b}$
Let $a+b-c=2z,\ b+c-a=2x,\ a+c-b=2y$. Without losing generality, assume $x \geq y \geq z$. Now, we need prove
$$\prod \left(x+y\right)^{(x+y)} \geq \prod \left(2z\right)^{(x+y)}$$
$$\prod \left(\frac{x+y}{2z}\right)^{(x+y)} \geq 1$$
$$\prod \left(\frac{x+y}{y}\right)^x \left(\frac{x+y}{x}\right)^y \geq 4^{x+y+z}$$
$$\prod \left(\frac{y+x}{2y}\right)^x \left(\frac{x+y}{2x}\right)^y \geq 1$$
Applying Cauchy theorem for left side, 
$$\prod \left(\frac{y+x}{2y}\right)^x \left(\frac{x+y}{2x}\right)^y \geq \prod \left(\frac{x}{y}\right)^{x/2} \left(\frac{y}{x}\right)^{y/2} = A$$
After simplifying $A$, we get
$$A = \left(\frac{x}{y}\right)^{\frac{x-y}{2}} \left(\frac{y}{z}\right)^{\frac{y-z}{2}} \left(\frac{x}{z}\right)^{\frac{x-z}{2}} \geq 1$$
Hence the given inequality holds iff $a=b=c$.
A: Another solution:
Prove: $$\left ( a+ b- c \right )^{a}\left ( b+ c- a \right )^{b}\left ( c+ a- b \right )^{c}\leq a^{a}b^{b}c^{c}$$
By weight power mean:
$$\sqrt[a+ b+ c]{\left ( \frac{a+ b- c}{a} \right )^{a}\left ( \frac{b+ c- a}{b} \right )^{b}\left ( \frac{c+ a- b}{c} \right )^{c}}\leq  \frac{1}{a+ b+ c}\left ( a. \frac{a+ b- c}{a}+ b. \frac{b+ c- a}{b}+ c. \frac{c+ a- b}{c} \right )= 1$$
i.e.
