let-a-b-c-IR-show-that-a-b-a-c-2-abc-a-b-c-

Question Number 71633 by mind is power last updated on 18/Oct/19

let a, b, c \in {IR}_{+}

show that (a + b) (a + c) ⩾ 2 \sqrt{abc (a + b + c)}

Answered by MJS last updated on 18/Oct/19

a > 0 \land b > 0 \land c > 0 \Rightarrow (a + b) (a + c) > 0 \land a b c (a + b + c) > 0

\Rightarrow we are allowed to square

{(a + b)}^{2} {(a + c)}^{2} ⩾ 4 a b c (a + b + c)

{(a + b)}^{2} {(a + c)}^{2} - 4 a b c (a + b + c) ⩾ 0

{(a - b)}^{2} c^{2} + 2 a (a - b) (a + b) c + a^{2} {(a + b)}^{2} ⩾ 0

{((a - b) c + a (a + b))}^{2} ⩾ 0

true \forall a, b, c \in R^{+}

Commented by mind is power last updated on 18/Oct/19

(a + b) (a + c) ⩾ 4 abc (a + b + c) y^{'} re methode worck nice sir

Commented by MJS last updated on 18/Oct/19

corrected, it was just a typo

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