Question

If a, b, c be the lengths of the sides of a triangle, then $(a+b+c{)}^3$ $\ge$ 27(a+b-c)(b+c-a)(c+a-b)

• A. True
• B. False

Answer 1

The correct option is A

True

Since the sum of 2 sides of a triangle is greater than the 3rd side, We have

a + b > c (or) a + b - c > 0

b + c > a (or) b + c - a > 0

c + a > b (or) a + c - b > 0

Applying A.M of 3 quantities $\ge$ G.M of 3 quantities

$\frac{(a+b-c)+(b+c-a)+(a+c-b)}{3}$ $\ge$$\left\{\right(a+b-c\left)\right(b+c-a\left)\right(a+c-b\left)\right\}$${}^{\frac{1}{3}}$

$\frac{a+b+c}{3}$ $\ge$ $\left[\right(a+b-c\left)\right(b+c-a\left)\right(a+c-b){]}^{\frac{1}{3}}$

$(a+b+c{)}^3$ $\ge$ 27 (a+b-c) (b+c-a) (c+a-b)