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Show that the product of two of the numbers $(65^{1000} - 8^{2001} + 3^{177}), (79^{1212} - 9^{2399} + 2^{2001})$ and $(24^{4493} - 5^{8192} + 7^{1777})$ is non-negative, without actually evaluating the numbers.

P.S. I have found by calculation that all the three numbers are positive, but that does not solve the problem of proving without calculation.

Thanks in advance.

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    What do you mean by "rigorous proof?" Are you just wanting a heuristic approach to solving the problem without the actual calculation?2011-11-09

4 Answers 4

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I think the trick is that the question just asks for a proof that there exist two of the numbers such that their product is non-negative. In principle there could be other products that were negative.

Now, if at least two of the numbers are non-negative, then their product is non-negative too.

If less than two of them are non-negative, there must be at least two negative numbers among them ...

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    I think Henning's interpretation is probably correct.2011-11-09
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Consider for instance $65^{1000} - 8^{2001} + 3^{177}$. Note that $65>64=8^2$ and so $65^{1000} > 8^{2000}$, but it'd be better if we had $65^{1000} > 8^{2001}$ because then $65^{1000} - 8^{2001} + 3^{177}$ would definitely be positive. So we need a finer estimate for $65^{1000}$. Here is one: $65^{1000} = (64+1)^{1000} = 64^{1000}+1000\cdot 64^{999}+\cdots > 64^{1000}+7\cdot 64 \cdot 64^{999} = 8 \cdot 64^{1000} = 8^{2001}$.

The other two numbers are positive in the same fashion but you need a different argument.

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    @Henning, I know... :-(2011-11-09
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Unless I did a mistake in the calculations, this should solve the last one.

$24^{4493}=2^{3*4493}*3^{4493}$

Now using

$2^7 \geq 5^3 \,;\, 3^3 \geq 5^2 \,,$

we get

$2^{3*4493}*3^{4493} \geq (2^7)^{1925}(3^3)^{1497} \geq 5^{1925*3+1497*2}=5^{8769}$

And here is the other

$(\frac{79}{81})^{20}= (1-\frac{2}{81})^{20} \geq 1-\frac{40}{81} \,.$

by Bernoulli

Thus

$(\frac{79}{81})^{100}\geq \frac{1}{2^5} \geq \frac{1}{79} \,.$

Thus

$79^{101} \geq 81^{100} \,.$

And hence

$79^{1212} \geq 81^{1200} $

The positivity of the second term is an immediate consequence of this....

P.S. Edit The last inequality also follows by this idea:

We show that

$24^{4493} \geq 25^{4096}$

$(\frac{24}{25})^{12} \geq 1-\frac{12}{25} \geq \frac{1}{2}$

$(\frac{24}{25})^{12*342}\geq \frac{1}{2^{342}} \geq{1}{24^{389}} \,.$