How to know if it is not possible to determine the sign of an expression? I would like to know if the following expression is always positive or negative:
$(1-t_c)[(1-\theta)+\theta c_1 (1-t)]-c_1(1-t)$
where:
$0<t_c<1$
$0<\theta<1$
$0<c_1<1$
$0<t<1$
The solution says the expression can be either positive or negative depending on the values of the variables.
I actually spent a few hours trying to arrive at an answer before giving up and looking at the solution. So, is there a way to quickly recognize that a given expression can be positive or negative depending on the values that the expression's variables take?
 A: No, there is no quick way. For algebraic expressions like the one you posted, you can expand them out and it would be just as hard as determining whether an arbitrary multivariate polynomial has zeros in a given region.
However, for your problem, set $A = (1 - t_c)[(1 - \theta)  + \theta c_1 (1 - t)]$ and $B = c_1(1 - t)$. Then $A > 0, B > 0$ and your expression is $A - B$. If you pick $t$ very close to $1$, then $B$ is very small and you can arrange for $A$ to be not small, so the expression can be positive. Similarly, you can arrange for $t_c$ to be very close to $1$, in which case $A$ is very small, and you can arrange for $B$ to be not small, so the expression can also be negative.
A: In general, no like Christopher A. Wong said. However in situations where all the variables are essentially linear (note even if you expand everything out you never get individual variables with powers higher than 1), you can often try setting the entire expression to $0$ and trying to find some solution. If you can, then it usually takes very little extra work to show it can be both positive and negative. If possible you want to get all of the same variable to one side or the other.
$(1-t_c)[(1-\theta)+\theta c_1 (1-t)]-c_1(1-t) = 0 \\
\Longleftrightarrow (1-t_c)(1-\theta)+(1-t_c)\theta c_1 (1-t)=c_1(1-t) \\
\Longleftrightarrow (1-t_c)(1-\theta) = c_1(1-t)[1-(1-t_c)\theta] \\
\Longleftrightarrow \dfrac{(1-t_c)(1-\theta)}{1-(1-t_c)\theta} = c_1(1-t)$
Now if we pick easy values to work with for the complicated LHS, say $t_c = \theta = \frac{1}{2}$, then the LHS simplifies down to $\frac{1}{3}$. We can easily get $\frac{1}{3}$ from the RHS by letting $c_1 = \frac{1}{2}$ and $t = \frac{1}{3}$. Thus we can pick values that make the whole expression $0$. To make it positive, leave everything alone except $c_1 = \frac{1}{100}$. To make it negative, leave everything along except $c_1 = \frac{99}{100}$.
