Average of square roots's sum vs. square root of an average I was watching a video on youtube about how colors work in computers, and found this statement:

"The average of two square roots is less than the square root of an
  average"

The link to the video where I found this: here
There's even an image in the video with the corresponding algebraic expression, even though in the image it's using a <= (less or equal) sign, instead of a < (less than) sign.
Could someone proof how this makes sense?
P.S. Sorry I don't include the expression in this question, I don't know how to.
 A: This is from the comment of André Nicolas, which I will just attempt to clarify.
Assume $x,y\ge0$.
We want $${[\sqrt x + \sqrt y]\over 2 } \le \sqrt{(x+y)\over 2}$$
We proceed with inequalities equivalent to the first and to each other:
$${[\sqrt x + \sqrt y]\over 2 } \le {\sqrt{(x+y)}\over \sqrt 2}$$
$${\sqrt x + \sqrt y } \le {2\sqrt{(x+y)}\over \sqrt 2}$$
$${\sqrt x + \sqrt y } \le {\sqrt2\sqrt{(x+y)}}$$
squaring
$$x+2\sqrt{xy}+y\le 2x+2y$$
$$x+y-2\sqrt{xy}\ge0$$
$${{[\sqrt x - \sqrt y]}}^2\ge0$$ which we know to be true. Thus, its equivalent inequality, the one we are trying to prove is also true.
A: Good question, I actually happened to have watched that video. Manipulate...
$${{\sqrt a+ \sqrt b} \over 2} \le \sqrt{{a+b} \over 2}$$
into something easier to work with. Square both sides, assuming that they are positive.
$${{{a+b} \over 4} + {\sqrt{ab} \over 2}} \le {{a+b} \over 2}$$
$${\sqrt{ab} \over 2} \le {{a+b} \over 4}$$
$${\sqrt{ab}} \le {{a+b} \over 2}$$
Now, subsitute $b=a+n$
$${\sqrt{a^2+an}} \le {{2a+n} \over 2}$$
assume the equality is true for some number $a$. Square both sides of the inequality and multiply by 4...
$${4a^2+4an} \le {4a^2+4an+n^2}$$
$$0 \le n^2$$ we already noted that both a and b are real and positive so this true. If you're still unclear, just do the steps in reverse and you'll get back to the original inequality.
A: This is no accident actually. In a broader context, the inequality is known as generalized mean inequality, check this wonderful wikipedia page:
https://en.wikipedia.org/wiki/Generalized_mean
In this special scenario, the result is a simply application of the fact that arithmetic mean (p = 1) is less than or equal to the quadratic mean (p=2). 
Cheers~
A: Many such inequalities between the average of a function and the function of the average follow directly from Jensen's inequality.
A: As Bill says, the square root is concave so just apply Jensen’s (usually stated for convex, so flip the inequality in the usual statement) for the expectation in the discrete case.
