A-2B-3C-4D-are-positive-numbers-forming-a-geometric-series-prov-that-A-3C-B-2D-gt-2-

Question Number 11956 by Mahmoud A.R last updated on 07/Apr/17

A, 2 B, 3 C, 4 D

a r e p o s i t i v e n u m b e r s f o r m i n g a

g e o m e t r i c s e r i e s

p r o v t h a t :

(A + 3 C) (B + 2 D) > 2

Commented by ajfour last updated on 07/Apr/17

i f A = \frac{1}{64}, 2 B = \frac{1}{32}, 3 C = \frac{1}{16}, 4 D = \frac{1}{8}

(A + 3 C) (B + 2 D) = \frac{5}{64} \times \frac{5}{64} > 2

!! ? ? ? ?

Commented by FilupS last updated on 07/Apr/17

A, B, C, D > 0

S e q u e n c e :

A, 2 B, 3 C, 4 D

∴ 2 B = n A \Rightarrow B = \frac{n A}{2}

3 C = n (2 B) = n^{2} A

4 D = n (3 C) = n^{3} A \Rightarrow 2 D = \frac{n^{3} A}{2}

(A + 3 C) (B + 2 D) = (A + n^{2} A) (\frac{n A}{2} + \frac{n^{3} A}{2})

= A^{2} (1 + n^{2}) (\frac{1}{2} (n + n^{3}))

= \frac{1}{2} {n A}^{2} (1 + n^{2}) (1 + n^{2})

= \frac{1}{2} {n A}^{2} {(1 + n^{2})}^{2}

{(1 + n^{2})}^{2} > 1

A^{2} > 0

let \frac{1}{2} A^{2} {(1 + n^{2})}^{2} = k, k > 0

= n k

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