If $a,b,c$ are sides of the $△ABC$ such that
${(1+\frac{b-c}{a})}^a\cdot {(1+\frac{c-a}{b})}^b\cdot {(1+\frac{a-b}{c})}^c\ge 1$, then triangle $△ABC$ must be

Given that ${\left(1+\frac{b-c}{a}\right)}^a\times {\left(1+\frac{c-a}{b}\right)}^b\times {\left(1+\frac{a-b}{c}\right)}^c\ge \mathrm{1....}(1)$

${\left(1+\frac{b-c}{a}\right)}^a=\left(1+\frac{b-c}{a}\right)\times \left(1+\frac{b-c}{a}\right)..........\left(1+\frac{b-c}{a}\right)(atimes)$

we can write similar equations wit the other two, there are a total of a+b+c terms

applying A.M$\ge$G.M we get

$\frac{\left(1+\frac{b-c}{a}\right)\times a+\left(1+\frac{c-a}{b}\right)\times b+\left(1+\frac{a-b}{c}\right)\times c}{a+b+c}\ge {\left[{\left(1+\frac{b-c}{a}\right)}^a\times {\left(1+\frac{c-a}{b}\right)}^b\times {\left(1+\frac{a-b}{c}\right)}^c\right]}^{\frac{1}{a+b+c}}$

The expression for $AM$ when simplified gives $1.$

$$\begin{gathered} \Rightarrow 1\ge {\left[{\left(1+\frac{b-c}{a}\right)}^a\times {\left(1+\frac{c-a}{b}\right)}^b\times {\left(1+\frac{a-b}{c}\right)}^c\right]}^{\frac{1}{a+b+c}} \\ \Rightarrow \left[{\left(1+\frac{b-c}{a}\right)}^a\times {\left(1+\frac{c-a}{b}\right)}^b\times {\left(1+\frac{a-b}{c}\right)}^c\right]\le \mathrm{1.........}(2) \\ \Rightarrow Fromthegivenequation(1)and(2), \\ \left[{\left(1+\frac{b-c}{a}\right)}^a\times {\left(1+\frac{c-a}{b}\right)}^b\times {\left(1+\frac{a-b}{c}\right)}^c\right]=1 \end{gathered}$$

as all terms can only be greater than or equal to 1 each should be equal to 1.

$$\begin{gathered} \Rightarrow {\left(1+\frac{b-c}{a}\right)}^a={\left(1+\frac{c-a}{b}\right)}^b={\left(1+\frac{a-b}{c}\right)}^c=1 \\ \Rightarrow \frac{b-c}{a}=\frac{c-a}{b}=\frac{a-b}{c}=0 \\ \Rightarrow a=b=c \end{gathered}$$

Hence the triangle is equilateral.