If a, b, c are lengths of the sides of a triangle, then (a + b + c)3 is

1) ≥ 3 (a + b – c) (b + c – a) (c + a – b)

2) ≥ 9 (a + b – c) (b + c – a) (c + a – b)

3) ≥ 27 (a + b – c) (b + c – a) (c + a – b)

4) none of these

Solution: (3) ≥ 27 (a + b – c) (b + c – a) (c + a – b)

a + b > c or a + b – c > 0

b + c > a or b + c – a > 0

c + a > b or c + a – b > 0

[(a + b – c) (b + c – a) (c + a – b)] ≥ [AM of 3 quantities ≥ GM of 3 quantities] [a + b + c] / 3 ≥ [(a + b – c) (b + c – a) (c + a – b)]

(a + b + c)3 ≥ 27 [(a + b – c) (b + c – a) (c + a – b)]

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