Question

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

• A. right angled
• B. isosceles
• C. obtuse
• D. equilateral

Answer 1

The correct option is D equilateral

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

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

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

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

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

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

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

$\Rightarrow {(1+\frac{b-c}{a})}^a={(1+\frac{c-a}{b})}^b={(1+\frac{a-b}{c})}^c=1\Rightarrow \frac{b-c}{a}=\frac{c-a}{b}=\frac{a-b}{c}=0\Rightarrow a=b=c$

Hence the triangle is equilateral.