Proving a variable less than a constant I've been working on a math problem for a while and I've gotten to the point where I'm trying to prove the following
If
$$A(x, y) = B(x, y) = c_1\hat{x}$$
$$C(x, y) = D(x, y) = c_2\hat{y}$$
$$E(x, y) = F(x, y) = c_1c_2\widehat{xy}$$
Then can I prove that $x<c_1$ and $y<c_2$?
This isn't a homework problem, and I've been working on this problem for a very long time, but I'm not a mathematician and I'm not sure how I would even begin to figure this out if it's true.
 A: I'm not sure what the original problem is, nor what all of the information is. Assuming that some information about $A,C$ and/or $E$ is either given or somehow required for this problem, let's see what information about $A,C$ and $E$ can be used to directly show your desired inequality. 
We're trying to prove that $x < c_{1}$ and $y < c_{2}$. Suppose first that $x < c_{1}$. Working backwards, let's express this into terms of $A,C,E$ through multiplying both sides by $AC=E$:
$$ A(x,y)C(x,y)x < c_{1}E(x,y).$$
Simple as that. Show that $A(x,y)C(x,y)x < c_{1}E(x,y)$, or any algebraic manipulation of it such as $\frac{A(x,y)C(x,y)}{c_{1}} < \frac{E(x,y)}{x}$, and $x < c_{1}$ directly follows. Similarly, $y < c_{2}$ is equivalent to 
$$A(x,y)C(x,y)y < c_{2}E(x,y).$$
Again, I'm not sure why $E$ is defined, since it is equal to $AC$, unless some information regarding $A,C,E$ is given. But these two conditions on $A,C,E$ prove your desired inequalities. Remember that given only the information you've provided in the question, there is no basis for the inequality.
