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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.

  • 1
    You might be interested in Chapter IV of Hardy-Littlewood-Polya's *Inequalities* which is more or less all devoted to this technique. By the way, I think that the jargon for $f$ is *deficit function*.2012-06-10

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