Average of square roots's sum vs. square root of an average

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.

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.

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~