Here's example:
Prove that $x-\frac{1}{x^2} \geq 0$ for every $x \ge 1$
I know that this can be done using elementary algebra, but in other cases it's not that simple. Can I prove this inequation positive for every $x \ge 1$ if I show that it's positive for the smallest value, that's 1 in this case and then that limit of the function $f(x) = x-\frac{1}{x^2}$ as x approaches infinitiy is infinity?
On
No, unfortunately you can't. Consider the function $y=x(x-1)(x-2)$ as a counterexample.
However, try looking at the derivative of the function as this will tell you how the function changes over time and hence if it will stay non-negative.
P.S. You may want to check your inequalities as currently your statement isn't true for $x=1$
No, you can't. There are functions that start and end positive but are negative somewhere in the middle. On the other hand, if you could show that your function is increasing...