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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Thread starter
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- #1

- Jan 26, 2012

- 4,055

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- #2

- Jan 26, 2012

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1) kaliprasad

2) MarkFL

3) jacobi

4) Sudharaka

Solution (from MarkFL):

$\displaystyle (\sqrt{1001}+\sqrt{999})^2\,?\,(2\sqrt{1000})^2$

$\displaystyle 1001+2\sqrt{1001\cdot999}+999\,?\,4000$

$\displaystyle \left(\sqrt{(1000+1)(1000-1)} \right)^2\,?\,1000^2$

$\displaystyle 1000^2-1\,?\,1000^2$

$\displaystyle -1\,?\,0$

$\displaystyle -1<0$

Thus, we may conclude:

$\displaystyle \sqrt{1001}+\sqrt{999}<2\sqrt{1000}$

Alternate solution (from Sudharaka):

&=&2(1000+\sqrt{10^6-1})~~~~~~~~~~~~(1)

\end{eqnarray}

Since, \(\sqrt{10^6-1}<\sqrt{10^6}=10^3\) we have,

\[2(1000+\sqrt{10^6-1})<4000=(2\sqrt{1000})^2~~~~~~~~~~~~(2)\]

By (1) and (2) we get,

\[(\sqrt{1001}+\sqrt{999})^2<(2\sqrt{1000})^2\]

\[\therefore \sqrt{1001}+\sqrt{999}<2\sqrt{1000}\]

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