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

$$A,2B,3C,4D$$ $$arepositivenumbersforminga$$ $$geometricseries$$ $$provthat:$$ $$(A+3C)(B+2D)>2$$

Commented byajfour last updated on 07/Apr/17

$$ifA=\frac{1}{64},2B=\frac{1}{32},3C=\frac{1}{16},4D=\frac{1}{8}$$ $$(A+3C)(B+2D)=\frac{5}{64}\times \frac{5}{64}>2$$ $$!!????$$

Commented byFilupS last updated on 07/Apr/17

$$A,B,C,D>0$$ $$$$ $$Sequence:$$ $$A,2B,3C,4D$$ $$\therefore 2B=nA\Rightarrow B=\frac{nA}{2}$$ $$3C=n\left(2B\right)=n^{2}A$$ $$4D=n\left(3C\right)=n^{3}A\Rightarrow 2D=\frac{n^{3}A}{2}$$ $$(A+3C)(B+2D)=(A+n^{2}A)(\frac{nA}{2}+\frac{n^{3}A}{2})$$ $$=A^2(1+n^2)\left(\frac{1}{2}(n+n^3)\right)$$ $$=\frac{1}{2}{nA}^2(1+n^2)(1+n^2)$$ $$=\frac{1}{2}{nA}^2{(1+n^2)}^2$$ $${(1+n^2)}^2>1$$ $$A^2>0$$ $$\mathrm{let}\frac{1}{2}{\mathrm{A}}^2{(1+n^2)}^2=k,k>0$$ $$=nk$$ $$$$ $$????$$

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