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I'm asked to square the inequality and use the Mean Value Theorem to prove that

$$\sqrt{1+x} < 1 + \frac{x}{2}$$ for $x>0$.

Unfortunately, I don't really understand why I would need such a theorem here in the first place. Squaring the inequality, we get:

$$1+x < 1 + x + \frac{x^2}{4}$$

and thus: $$0 < \frac{x^2}{4}$$

and therefore: $$x^2 > 0$$

which obviously holds iff $x \neq 0$. Therefore, this especially holds for $x > 0$, q.e.d.

Sadly, the problem explicitly asks to use the Mean Value Theorem. Can anyone explain why I "need" and how I would apply the Mean Value Theorem here? What's the deal?

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  • $\begingroup$ Can you give the exact problem statement? $\endgroup$ – user61527 Jun 5 '14 at 2:27
  • $\begingroup$ Show (by squaring and applying the Mean Value Theorem) that for $x > 0: \sqrt{1 + x} < 1 + \frac{x}{2}$. $\endgroup$ – chiru Jun 5 '14 at 2:30
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    $\begingroup$ I would interpret it as asking to show the result twice (i.e., to two different proofs), once by squaring, and the second by applying the MVT. $\endgroup$ – Clement C. Jun 5 '14 at 2:32
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    $\begingroup$ I point out that you're wrong in saying that $x^2>0$ holds iff $x>0$. In fact, it holds for all real $x$ with the single exception of $0$. In addition, your chain of implications goes in exactly the wrong direction. You are assuming the result and "proving" the hypothesis. $\endgroup$ – MPW Jun 5 '14 at 2:34
  • $\begingroup$ @MPW You're right, thanks! Edited and hopefully corrected. :) $\endgroup$ – chiru Jun 5 '14 at 2:36
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Let $f(x) = \sqrt{1+x}$ for $x > 0$. On the interval $(0,x)$:

$f(x) - f(0) = f'(c)(x - 0)$ for some $c \in (0,x)$. This gives:

$\sqrt{1+x} - 1 = \dfrac{x}{2\sqrt{1+c}} < \dfrac{x}{2}$ since $1 + c > 1$

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  • $\begingroup$ I see, thanks a lot! Could you leave a comment about the criteria that you have for choosing a good $f(x)?$ I might as well have chosen $f(x) := 1 + \frac{x}{2}$, couldn't I have? Is either side of the inequality going to lead me somewhere? $\endgroup$ – chiru Jun 5 '14 at 2:53
  • $\begingroup$ Observe that: $1 = (1+0)^{1/2}$ $\endgroup$ – DeepSea Jun 5 '14 at 3:11
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Hint. First write down carefully and accurately the Mean Value Theorem. Then see if you can fill in missing steps and provide reasons for the following: $$\frac{\sqrt{1+x}-\sqrt{1+0}}{x-0}=\frac{1}{2\sqrt{1+c}}<\frac{1}{2}$$ where $0<c<x$. If you can do this then the rest of the problem should be easy.

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