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The clue given by the text is to "use the fact that $\sqrt{x}$ is increasing."

I was able to get the correct answer here by squaring both expressions. But I don't think I made use of the text-prided clue, so am wondering if perhaps I'm missing the "deeper" lesson.

What do you think the author is hinting at?

Thanks!

drzaius7
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    I wonder if the author wanted to use that $\sqrt{x}$ is concave rather than increasing - that condition would suffice... – Milo Brandt Apr 20 '15 at 00:37
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    Not an answer to your question, but I think a "deeper" lesson here could be had by proving the inequality using only the fact that $\sqrt{\cdot}$ is concave (and that $\frac{3+5}{2} = \frac{2+6}{2}$). –  Apr 20 '15 at 00:37
  • ... as @Meelo said. –  Apr 20 '15 at 00:37
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    I have to say, even though one only needs the fact that $\sqrt{x}$ is increasing, that's not much of a hint. After all, $x^2$ is increasing, and $3^2+5^2$ is certainly less than $2^2+6^2$. A better hint would have been to square the two expressions. I suppose the fact that $3+5 = 2+6$ suggests that squaring would help, but if you saw that, you probably wouldn't need a hint on the other. – Brian Tung Apr 20 '15 at 00:46

8 Answers8

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Hint: Easier to show that $$\sqrt{6}-\sqrt{5}<\sqrt{3}-\sqrt{2}$$

Thomas Andrews
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  • It seems odd that this answer of mine is the highest-voted answer, since the more straightforward answer is also perhaps more elegant - squaring both sides. – Thomas Andrews Apr 22 '15 at 19:19
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The square-rooting operation on positive numbers is a strictly increasing mapping. That means that if $x^2 < y^2$, you can conclude $x < y$. This is the definition of strictly increasing:

$$a < b \implies f(a) < f(b)$$

By comparing the squares of the numbers instead of the numbers, you are using the prompted clue.

GFauxPas
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Since $x\mapsto\sqrt{x}$ is an increasing function we have the following \begin{align*} 2\sqrt{15}&>2\sqrt{12}\\ 8+2\sqrt{15}&>8+2\sqrt{12}\\ 3+2\sqrt{15}+5&>2+2\sqrt{12}+6\\ (\sqrt{3}+\sqrt{5})^2&>(\sqrt{2}+\sqrt{6})^2 \end{align*} Therefore, $\sqrt{3}+\sqrt{5}>\sqrt{2}+\sqrt{6}$.

Ángel Mario Gallegos
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Square both sides. You have $$ \sqrt{3} + \sqrt{5} \mathrel{?} \sqrt{2} + \sqrt{6}.$$ Squaring you get $$ 8 + 2\sqrt{15} \mathrel{?} 8 + 2\sqrt{12};$$ paring you get $$ \sqrt{15} \mathrel{?} \sqrt{12}.$$

Now it is entirely clear.

ncmathsadist
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Square the two numbers and we're left to compare $8+2\sqrt{15}$ and $8+2\sqrt{12}$. Because $\sqrt{x}$ is increasing we have

$$\sqrt{12}\lt \sqrt{15}$$

$$8+2\sqrt{12}\lt 8+2\sqrt{15}$$

And therefore

$$\sqrt{2}+\sqrt{6}\lt \sqrt{3}+\sqrt{5}$$

marwalix
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The square root function is increasing and concave down (meaning it increases slower and slower as $x$ gets larger). Thus $\sqrt{3}-\sqrt{2}>\sqrt{6}-\sqrt{5}$.

Rearranging gives $\sqrt{3}+\sqrt{5}>\sqrt{2}+\sqrt{6}$.

paw88789
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It's not just that $\surd x$ is increasing, it's the way that it is increasing.

Although the curve, $y=\surd x$, is increasing monotonically towards the positive infinite as $x$ trends towards the positive infinite, the slope of the curve is decreasing monotonically towards zero. It is concave.

Hence the difference between $\surd 3$ and $\surd 2$ will be larger than the difference between $\surd 6$ and $\surd 5$.

$$\because \surd 3 - \surd 2 > \surd 6-\surd 5 \\[2ex] \therefore \surd 3 + \surd 5 > \surd 2+ \surd 6$$

Graham Kemp
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  • In general you are right but we have a very concrete question here. In this special casse, square both sides and you reduce the question to $\sqrt{15} \stackrel{?}{>} \sqrt{12}$ which can be answered using the given hint. – Alfe May 16 '17 at 10:59
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Putting Graham’s answer into a picture, sort of, which is greater, the combined heights of the green segments or the combined heights of the red segments one unit away in each direction?

enter image description here

Steve Kass
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