I have noticed a few answers involving inequalities proven by creating a function comprised of the terms in the inequality. I hadn't seen this before and was wanting to know more about how to use it.
From what I have seen, it goes something like:
I don't know if this is a good example, but say we wanted to prove: $\sqrt{x+4} - 2 < \frac{x}{4}$
If I defined $f$ to be $f(x) = \frac{x}{4} + 2 -\sqrt{x+4}~$ (which would have a domain of $[-4, \infty) \in \mathbb{R}$), then if I can show that it is always positive then this will prove my inequality right?
$f'(x) = \frac{1}{4} - \frac{1}{2\sqrt{x+4}}$
$\sqrt{x+4} = 2$
$x = 0$
So, $x = 0$ is a critical point.
$f''(0) = \frac{1}{32} > 0$
Therefore, $ 0 $ is a minimum.
$f(0) = 0$
So, $0$ is the minimum value of the function, i.e., $f(x) \geq 0$ for all defined $x$.
Is this all correct? What about the original inequality, how do I account for all $x$, if my function only proved it for $[-4, \infty)$?
Basically, I just wanted to know about the strategy and everything that I can use it for.