<s>[INST] 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
 [/INST] 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.
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