Which is larger, $\sqrt{3} + \sqrt{5}$ or $\sqrt{2} + \sqrt{6}$? 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!
 A: 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.
A: 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, $\sqrt{3}+\sqrt{5}>\sqrt{2}+\sqrt{6}$.
A: 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.
A: Hint: Easier to show that $$\sqrt{6}-\sqrt{5}<\sqrt{3}-\sqrt{2}$$
A: 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 
$$\sqrt{2}+\sqrt{6}\lt \sqrt{3}+\sqrt{5}$$
A: 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 $\sqrt{3}+\sqrt{5}>\sqrt{2}+\sqrt{6}$.
A: 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 $\surd 5$.
$$\because \surd 3 - \surd 2 > \surd 6-\surd 5 \\[2ex] \therefore \surd 3 + \surd 5 > \surd 2+ \surd 6$$
A: 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?

