A , 2B ,3C ,4D are positive numbers forming a geometric series prov that : (A + 3C) (B + 2D)


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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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