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Problem of the week #60 - May 20th, 2013

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Jameson

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Jan 26, 2012
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Without using a calculator determine which is larger: \(\displaystyle \sqrt{1001}+\sqrt{999}\) or \(\displaystyle 2\sqrt{1000}\).
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) kaliprasad
2) MarkFL
3) jacobi
4) Sudharaka

Solution (from MarkFL):
$\displaystyle \sqrt{1001}+\sqrt{999}\,?\,2\sqrt{1000}$

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

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

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

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

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

$\displaystyle -1<0$

Thus, we may conclude:

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


Alternate solution (from Sudharaka):
\begin{eqnarray} (\sqrt{1001}+\sqrt{999})^2 &=& 2(1000+\sqrt{1001\times 999})\\

&=&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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