Question

If a+b+c=13 then , what is the HCF of (a-3)(b-2)(c+1)

Answer 1

Let a-3, b-2 and c+1 be termed as x, y, z respectively.Then by the inequality, we know that

$(\frac{x + y + z}{3} ) \geqslant (xyz) \frac{1}{3} \\ = > \frac{(a - 3) + (b - 2) + (c + 1)}{3} \geqslant (a - 3)(b - 2)(c +1) \\ = > \frac{(a + b + c - 4)}{3} \geqslant (a - 3)(b - 2)(c + 1) {1 \3}^{} \\ = > \frac{(13 - 4)}{3} \geqslant (a - 3)(b - 2)(c + 1) {1 \3}^{} \\ = > 3 \geqslant (a - 3)(b - 2)(c + 1) {1 \3}^{} \\ = > (3 ) {}^{3} \geqslant (a - 3)(b - 2)(c + 1) \\ = > (a - 3)(b - 2)(c + 1) \leqslant 27$

•Thus The Maximum Value of (a-3)(b-2)(c+1) is 27

NOBITA221✔