Proof without calculation 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.
 A: 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 ...
A: 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.
A: 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
$$24^{4493} \geq 25^{4096}$$
$$(\frac{24}{25})^{12} \geq 1-\frac{12}{25} \geq \frac{1}{2}$$
$$(\frac{24}{25})^{12*342}\geq \frac{1}{2^{342}} \geq{1}{24^{389}} \,.$$
A: Building on @henning-makholm this is a question in the 8th Ed textbook by Rosen. I solved it as follows:
Let $P(x,y)$ represent the predicate $xy > 0$, and the domain be $a,b,c$ represented by $(65^{1000}−8^{2001}+3^{177}),(79^{1212}−9^{2399}+2^{2001}),(24^{4493}−5^{8192}+7^{1777})$ respectively.


*

*We can conclude that $abc \ne 0$

*Case i) At least 2 of $a,b,c$ are non-negative

*Case ii) At least 2 of $a,b,c$ are negative


We know that the sum of two integers with the same sign will result in a non-negative number.
$\therefore \forall x \exists yP(x,y)$ such that $xy>0$. 
This is a nonconstructive proof. 
