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Show that, if $a$ and $h$ are positive numbers, $h < a^2$, then $$\sqrt{a^2 + h}-a < \frac{h}{2a} < a - \sqrt{a^2 - h}$$

I've been working on this problem for about 2 hours now, but I've made no progress. I'm not looking for an answer, but I just need some help to get me started since we didn't practice inequalities this complex in highschool. Thanks.

All I can tell is that we're supposed to take the square root of an expression at some point since one inequality ( h < a^2 ) becomes two.

Edit : Thank you guys for the replies, but I'd appreciate only hints in the future (like Robert Israel) so that I can learn. Regardless, I found a different way to do it, so it's cool :)

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More hints:

I'll use "$:$" as a placeholder for "$<$" or "$>$" since we don't know the nature of each inequality yet.

$$\begin{aligned}&\frac{h}{2a}:a-\sqrt{a^2-h}\\ &\Rightarrow \sqrt{a^2-h}:a-\frac{h}{2a}\\ &\Rightarrow \color{blue}{a^2-h}:\color{blue}{a^2-h}+\color{red}{\frac{h^2}{4a^2}}\\ \end{aligned}\\ $$


$$\begin{aligned}&\sqrt{a^2+h}-a:\frac{h}{2a}\\ &\Rightarrow \sqrt{a^2+h}:a+\frac{h}{2a}\\ &\Rightarrow \color{blue}{a^2+h}:\color{blue}{a^2+h}+\color{red}{\frac{h^2}{4a^2}}\\ \end{aligned}\\ $$

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Hints:

  1. Consider the two $<$'s separately.
  2. $A > B$ is equivalent to $A + C > B + C$.
  3. If $A > 0$ and $B > 0$, $\sqrt{A} < B$ is equivalent to $A < B^2$.
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  • $\begingroup$ Thank you, but I don't really get the first hint.. $\endgroup$ – user440261 Sep 9 '17 at 1:16
  • $\begingroup$ @Saad I believe what Robert means is 'first, think about $(h/2a< \ldots)$ and then, think about $(\ldots< h/2a)$". $\endgroup$ – Jam Sep 9 '17 at 1:22
  • $\begingroup$ So I should try working backwords to see how the expression arrived in the first place? $\endgroup$ – user440261 Sep 9 '17 at 1:23
  • $\begingroup$ The left side is equivalent to $\sqrt{a^2+h} < a + \dfrac{h}{2a}$ and the right side, $\sqrt{a^2 - h} < a - \dfrac{h}{2a}$ go from there. $\endgroup$ – steven gregory Sep 9 '17 at 1:35
  • $\begingroup$ Thank you. I think I got it. $\endgroup$ – user440261 Sep 9 '17 at 1:50
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Note that $$a-\sqrt{a^2-h}=\frac{\left(a-\sqrt{a^2-h}\right)\left(a+\sqrt{a^2-h}\right)}{a+\sqrt{a^2-h}}=\frac{h}{a+\sqrt{a^2-h}}.$$ So $$\frac{h}{2a} < a-\sqrt{a^2-h} \iff \frac{h}{2a} < \frac{h}{a+\sqrt{a^2-h}} \iff \sqrt{a^2-h}<a,$$ which is true.

Similarly, the left inequality can be proved.

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You should be able to prove that for $a,b \ge 0$ then $a< b \iff a^2 < b^2$. (prove that now.)

So $ \sqrt{a^2 + h} - a < \frac h{2a}\iff$

$\sqrt{a^2 + h} < \frac h{2a} + a\iff$

$a^2 + h < (\frac h{2a} + a)^2 = \frac {h^2}{4a^2} + h + a^2 \iff $

$0 < \frac{h^2}{4a^2}$ which it is.

Do the same to prove $\frac h{2a} < a - \sqrt{a^2 -h} \iff$

$\frac h{2a} - a < - \sqrt{a^2 -h}<0 \iff$

$(\frac h{2a})^2 - h + a^2 > a^2 - h \iff$

$\frac {h^2}{4a^2} > 0$ which it is.

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In questions like this, I think one variable is better than two. With two variables you are trying to prove $$\sqrt{a^2 + h}-a \stackrel?< \frac{h}{2a} \stackrel?< a - \sqrt{a^2 - h}.$$

Notice what happens when you divide everything by $a$ (which doesn't affect the truth of any inequalities, since $a$ is positive): $$\sqrt{1 + \frac{h}{a^2}}-1 \stackrel?< \frac{h}{2a^2} \stackrel?< 1 - \sqrt{1 - \frac{h}{a^2}}.$$

Set $x=\frac{h}{a^2},$ and what we have to prove is $$\sqrt{1 + x}-1 \stackrel?< \tfrac12x \stackrel?< 1 - \sqrt{1 - x}.$$

Since $h < a^2$ and since $a$ and $h$ are both positive, we know that $0<x<1.$ You can use those facts, along with some algebraic manipulation, to put each of the two inequalities you have to prove into an easily provable form. This somehow seems easier to me when there is only one variable on each side of each inequality rather than the way the inequalities were first presented.

Once you have two inequalities proved in a simplified form, you just need to work your way backwards, proving each unproved inequality above from the ones below it.

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