Question

If a, b, c be the lengths of the sides of a triangle and (a + b + c)³ ≥ k (a + b – c) (b + c – a) (c + a – b), then k equals

(a) 1 

(b) 3 

(c) 8 

(d) 27

Answer 1

(d) 27

By triangle inequality, we have sum of lengths of two sides of a Δ > length of third side 

⇒ a + b > c, b + c > a, c + a > b 

⇒ a + b – c > 0, b + c – a > 0, c + a – b > 0 

All being positive quantities, we apply AM > GM

⇒ \(\frac{(a+b-c)+(b+c-a)+(c+a-b)}{3}\) ≥ [(a+b-c)(b+c-a)(c+a-b)]^\(\frac13\)

⇒ \(\big(\frac{a+b+c}{3}\big)^3\) ≥ [(a+b-c)(b+c-a)(c+a-b)]

(a+b+c)3 ≥ 27(a+b-c)(b+c-a)(c+a-b)

⇒ K = 27.