<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] 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 $a=b=c$.
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