QQuestion
If (b + c – a)x = (c + a – b)y = (a + b – c)z = 2, then prove that (1/x + 1/y)(1/y + 1/z)(1/z + 1/x) = abc.
✓Answer
Given: (b + c – a)x = (c + a – b)y = (a + b – c)z = 2
x = 2\over \text{b + c - a} or 1/x = \text{b + c - a}\over 2
y = 2\over \text{c + a - b} or, 1/y = \text{c + a - b}\over 2
z = 2\over \text{a + b - c} or, 1/z = \text{a + b - c}\over 2
Calculating (1/x + 1/y):
(1/x + 1/y) = \text{b + c - a}\over 2 + \text{c + a - b}\over 2
= \text{b + c - a + c + a - b}\over 2
= c
Calculating (1/y + 1/z):
(1/y + 1/z) = \text{c + a - b}\over 2 + \text{a + b - c}\over 2
= \text{c + a - b + a + b - c}\over 2
= a
Calculating (1/z + 1/x):
(1/z + 1/x) = \text{a + b - c}\over 2 + \text{b + c - a}\over 2
= \text{a + b - c + b + c - a}\over 2
= b
LHS: (1/x + 1/y)(1/y + 1/z)(1/z + 1/x)
= c × b × a
= abc RHS
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