Which is larger, $\sqrt{3} + \sqrt{5}$ or $\sqrt{2} + \sqrt{6}$?

Question

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!

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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

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

Answer 5

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.

Answer 6

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

Answer 7

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 $\surd 5$.

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

Answer 8

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?