Tricks to solve inequalities I am wondering if there are some tricks to solve inequalities which are not manageable analytically. For example consider the inequality (say we restrict on positive $x$):
$\displaystyle
\frac {\text e^{-x^2}}{x^2}<1
$
Let's say I want to find a good lower bound for $x$ which solves this inequality.
But I can not solve it analytically to end up with a statement like: For all $x>...$ the inequality is fulfilled.
Unfortunately I just know the two usual tricks like


*

*estimate $\frac {\text e^{-x^2}}{x^2}$ against some other function which is manageable. For example I know that $\frac {\text e^{-x^2}}{x^2}<1/x^2$ and $1/x^2<1$ for all $x>1$. So $x>1$ gives me a lower bound for $x$. 

*try some value and use the fact that $\frac {\text e^{-x^2}}{x^2}$ is strictly decreasing. So I find that $x=0.8$ solves the inequality which gives me the slightly improved bound: $x>0.8$


Are there some more tricks? Some general treatments how to deal with inequalities where I have a precise expression of $x$ but can not solve it for $x>...$ analytically?
 A: In some cases, tables of function values work just fine. For example, the cumulative distribution function for Gaussian law has no analytical expression, yet its quantiles are important, so people just calculated its values with a sufficient precision ("sufficient" varies form task to task).
In other cases, CAS software can be useful.If you're trying to look at the behaviour of some function (take your problem, for instance), you can easily build a plot in, say, Mathematica, and even get a solution to your problem with a sufficient precision (yet it won't give the exact bound, of course).
A: Replacing $x$ by $x^2$, you are trying to find out when $e^{-x} - x < 0$. You can use Newton's method to approximate the zeroes of $e^{-x} - x$. Namely, start at $x = 0.9$ or some other value near a zero of $e^{-x} - x$ and iterate Newton's method aka Newton-Raphson as many times as you want. Since the derivative of $e^{-x} - x$ is negative near the zero in question, $e^{-x} - x$ is negative to the right of this zero, and you have a good approximation of the set you are looking for. A similar method can be used for other inequalities lacking an analytic solution. 
A: [ This is not a complete answer, but I think it illustrates a possible technique for this class of problems.]
You can start by numerically solving the problem to find the point of equality.
In this case, using Excel's goal seek, I find $x = 0.753186$
Now, you want to show this result analytically. We can see from the numerical solution that the value we desire is close to $3/4$.
So we can set: 
$x= {3\over4}+\alpha$
We then have:
$e^{-x^2}<x^2$ 
$e^{({3\over4}+\alpha)^2}<{({3\over4}+\alpha)^2}$
You can then expand the power series. You need to try to show that $\alpha$ is small in concert with placing bounds of the higher terms in the expansion.
