Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 118810 by greg_ed last updated on 19/Oct/20

$$\sqrt{2003}+\sqrt{2005}<2\sqrt{2004}???$$

Answered by Olaf last updated on 19/Oct/20

$$$$ $$x=\sqrt{2003}$$ $$x=\sqrt{2004-1}$$ $$x=\sqrt{2004}\sqrt{1-\frac{1}{2004}}$$ $$x=\sqrt{2004}(1-\frac{1}{2}.\frac{1}{2004}-\frac{1}{8}.\frac{1}{{2004}^2}...)$$ $$$$ $$y=\sqrt{2005}$$ $$y=\sqrt{2004+1}$$ $$y=\sqrt{2004}\sqrt{1+\frac{1}{2004}}$$ $$y=\sqrt{2004}(1+\frac{1}{2}.\frac{1}{2004}-\frac{1}{8}.\frac{1}{{2004}^2}...)$$ $$$$ $$x+y=2\sqrt{2004}-\frac{1}{4.{2004}^{3/2}}...⩽2\sqrt{2004}$$

Commented bygreg_ed last updated on 20/Oct/20

$$\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{help}!$$

Answered by MJS_new last updated on 19/Oct/20

$$\sqrt{a-1}+\sqrt{a+1}<2\sqrt{a}$$ $$\mathrm{defined}\mathrm{for}a⩾1$$ $$\mathrm{both}\mathrm{sides}>0\Rightarrow \mathrm{we}\mathrm{are}\mathrm{allowed}\mathrm{to}\mathrm{square}$$ $$a-1+2\sqrt{a-1}\sqrt{a+1}+a+1<4a$$ $$2a+2\sqrt{a-1}\sqrt{a+1}<4a$$ $$\sqrt{a-1}\sqrt{a+1}<a$$ $$\mathrm{squaring}\mathrm{again}$$ $$a^2-1<a^2$$ $$\mathrm{true}\mathrm{\forall }a⩾1$$

Commented bygreg_ed last updated on 20/Oct/20

$$\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{help}!$$

Answered by 1549442205PVT last updated on 20/Oct/20

$$\mathrm{Apply}\mathrm{the}\mathrm{inequality}(\mathrm{x}+\mathrm{y})<\sqrt{2({\mathrm{x}}^2+{\mathrm{y}}^2)}$$ $$\mathrm{for}\mathrm{\forall }\mathrm{x},\mathrm{y}>0\mathrm{x}\ne \mathrm{y}\mathrm{we}\mathrm{have}$$ $$\sqrt{2003}+\sqrt{2005}<\sqrt{2(2003+2005)}$$ $$=\sqrt{4.2004}=2\sqrt{2004}$$

Commented bygreg_ed last updated on 20/Oct/20

$$\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{help}!$$

Commented by1549442205PVT last updated on 25/Oct/20

$$\mathrm{You}\mathrm{are}\mathrm{welcome}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com