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Question Number 209234 by Tawa11 last updated on 04/Jul/24

$$\mathrm{Arrange}\mathrm{in}\mathrm{descending}\mathrm{order}:$$$$\sqrt{5}-\sqrt{2},\sqrt{7}-\sqrt{5},\sqrt{13}-\sqrt{11},\sqrt{19}-\sqrt{17}$$

Answered by A5T last updated on 04/Jul/24

$$Samearrangement:$$$$\sqrt{5}-\sqrt{2}(>\sqrt{5}-\sqrt{3})>\sqrt{7}-\sqrt{5}>\sqrt{13}-\sqrt{11}>\sqrt{19}-\sqrt{17}$$$$\sqrt{x+2}-\sqrt{x}isdecreasingas{x}^{+}increases$$$$\sqrt{x+2}-\sqrt{x}\stackrel{?}{>}\sqrt{x+3}-\sqrt{x+1}$$$$\iff x+2-x-2\sqrt{x(x+2)}>x+3-x-1-2\sqrt{(x+1)(x+3)}$$$$\iff x(x+2)<(x+1)(x+3)whichistrue.$$

Commented by Tawa11 last updated on 05/Jul/24

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Answered by BaliramKumar last updated on 07/Jul/24

$$\sqrt{5}-\sqrt{2}=\frac{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}{(\sqrt{5}+\sqrt{2})}=\frac{2}{(\sqrt{5}+\sqrt{2})}......\left(\mathrm{i}\right)$$$$\sqrt{7}-\sqrt{5}=\frac{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7}+\sqrt{5})}=\frac{2}{(\sqrt{7}+\sqrt{5})}......\left(\mathrm{ii}\right)$$$$\sqrt{13}-\sqrt{11}=\frac{(\sqrt{13}-\sqrt{11})(\sqrt{13}+\sqrt{11})}{(\sqrt{13}+\sqrt{11})}=\frac{2}{(\sqrt{13}+\sqrt{11})}...\left(\mathrm{iii}\right)$$$$\sqrt{19}-\sqrt{17}=\frac{(\sqrt{19}-\sqrt{17})(\sqrt{19}+\sqrt{17})}{(\sqrt{19}+\sqrt{17})}=\frac{2}{(\sqrt{19}+\sqrt{17})}....\left(\mathrm{iv}\right)$$$$\left(\mathrm{i}\right)>\left(\mathrm{ii}\right)>\left(\mathrm{iii}\right)>\left(\mathrm{iv}\right)$$$$$$

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