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 &lt;= (less or equal) sign, instead of a &lt; (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.
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.
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.