\((b + c - a)x = (c + a - b)y\) \( = (a + b - c)z = 2\)
\(\therefore \cfrac{1}{(b + c - a)x} = \cfrac{1}{(c + a - b)y}\)
\( = \cfrac{1}{(a + b - c)z} = \cfrac{1}{2}\)
\(\therefore \cfrac{1}{x}=\cfrac{b+c-a}{2}\)
\(\cfrac{1}{y}=\cfrac{c+a-b}{2}\)
এবং \(\cfrac{1}{z}=\cfrac{a+b-c}{2}\)
\(\therefore \cfrac{1}{x}+\cfrac{1}{y}=\cfrac{b+c-a}{2}+\cfrac{c+a-b}{2}\)
\(=\cfrac{b+c-a+c+a-b}{2}=\cfrac{2c}{2}=c\)
\(\cfrac{1}{y}+\cfrac{1}{z}=\cfrac{c+a-b}{2}+\cfrac{a+b-c}{2}\)
\(=\cfrac{c+a-b+a+b-c}{2}=\cfrac{2a}{2}=a\)
এবং
\(\cfrac{1}{z}+\cfrac{1}{x}=\cfrac{a+b-c}{2}+\cfrac{b+c-a}{2}\)
\(=\cfrac{a+b-c+b+c-a}{2}=\cfrac{2b}{2}=b\)
\(\therefore \left(\cfrac{1}{x}+\cfrac{1}{y}\right)\left(\cfrac{1}{y}+\cfrac{1}{z}\right)\left(\cfrac{1}{z}+\cfrac{1}{x}\right)=abc\)
(প্রমাণিত)