If a, b, c be the lengths of the sides of a triangle and (a + b + c)3

1.	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» (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.

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»

(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.