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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 is the goal of your question? What type of solution do you want? Intuitively, in any of the given parentheses the largest base is substantially larger than the others and with large exponents so one would think that the answer would be positive. –  analysisj Nov 9 '11 at 1:09
    
No, not intuitively, a more rigorous proof would be appreciated. And the smaller bases have larger powers, it's not so obvious that the numbers are positive. –  Shah Tamzid Nov 9 '11 at 1:19
    
What do you mean by "rigorous proof?" Are you just wanting a heuristic approach to solving the problem without the actual calculation? –  analysisj Nov 9 '11 at 1:34
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3 Answers

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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If two of them are positive but the third is negative then the product is negative. –  analysisj Nov 9 '11 at 1:21
    
It says "the product of two of the numbers ...". –  Henning Makholm Nov 9 '11 at 1:21
    
@analysisjb: Henning means that if at least two of them are non-negative, then there are two whose product is non-negative. He’s not talking about the product of all three. –  Brian M. Scott Nov 9 '11 at 1:23
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oh yeah! I get it! Thanks a lot! –  Shah Tamzid Nov 9 '11 at 1:24
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I think Henning's interpretation is probably correct. –  Grumpy Parsnip Nov 9 '11 at 1:53
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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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Unfortunately it doesn't work as easily for the other two expressions because $79$ is less than $9^2$ and similarly $24<5^2$. –  Henning Makholm Nov 9 '11 at 1:37
    
@Henning, I know... :-( –  lhf Nov 9 '11 at 1:40
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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

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

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

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

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