a-b-c-R-prove-that-a-b-c-3-3-abc-

Question Number 176399 by Matica last updated on 18/Sep/22

a, b, c \in R_{+}^{*} p r o v e t h a t a + b + c ⩾ 3^{3} \sqrt{a b c}

Commented by Matica last updated on 18/Sep/22

P l e a s e p r o v e t h a t A M ⩾ G M

Answered by Frix last updated on 18/Sep/22

for 2 numbers ⩾ 0

\frac{a + b}{2} ⩾ \sqrt{a b} both sides ⩾ 0 \Rightarrow we are allowed to square

\frac{{(a + b)}^{2}}{4} ⩾ a b

{(a + b)}^{2} ⩾ 4 a b

a^{2} + 2 a b + b^{2} ⩾ 4 a b

a^{2} - 2 a b + b^{2} ⩾ 0

{(a - b)}^{2} ⩾ 0 true

Answered by Frix last updated on 18/Sep/22

for 3 numbers ⩾ 0

\frac{a + b + c}{3} ⩾ \sqrt[3]{a b c}

\frac{{(a + b + c)}^{3}}{27} ⩾ a b c

{(a + b + c)}^{3} ⩾ 27 a b c

let a ⩽ b \land a ⩽ c \land p, q ⩾ 0

b = a + p \land c = a + q

{(3 a + p + q)}^{3} ⩾ 27 a (a + p) (a + q)

{(3 a + p + q)}^{3} - 27 a (a + p) (a + q) ⩾ 0

9 {a p}^{2} - 9 a p q + 9 {a q}^{2} + p^{3} + 3 p^{2} q + 3 {p q}^{2} + q^{3} ⩾ 0

9 a (p^{2} - p q + q^{2}) + {(p + q)}^{3} ⩾ 0 true

a ⩾ 0 \land {(p + q)}^{3} ⩾ 0

p^{2} - p q + q^{2} = {(p - q)}^{2} + p q ⩾ 0

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