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

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#### Jameson

Staff member
Without using a calculator determine which is larger: $$\displaystyle \sqrt{1001}+\sqrt{999}$$ or $$\displaystyle 2\sqrt{1000}$$.
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#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

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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