Question

If a, b, c are the lengths of the sides of a non-equilateral triangle, then, \(\frac1{a+b-c}+\frac1{b+c-a}+\frac1{c+a-b}\) is

(a) > \(\frac9{a+b+c}\)

(b) \(\frac1a+\frac1b+\frac1c\)

(c) Both (a) and (b) 

(d) None of these

Answer 1

(c) Both (a) and (b).

Since a, b, c are the sides of a non-equilateral triangle, a > 0, b > 0, c > 0. Also by triangle inequality. 

a + b > c ⇒ a + b – c > 0 

b + c > a ⇒ b + c – a > 0 

c + a > b ⇒ c + a – b > 0

∴ \(\frac{(a+b-c)^{-1}(b+c-a)^{-1}(c+a-b)^{-1}}{3}\) > \(\bigg\{\frac{(a+b-c)(b+c-a)(c+a-b)}{3}\bigg\}^{-1}\)

[∵ AM of the mth powers of n positive quantities is greater than the mth power of their AM if m < 0 or m > 1]

Hence (a) and (b) are both correct option.