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.

  • 3
    $\begingroup$ Not always $\lt$, if $x=y$ we have equality. We want to show that $\frac{\sqrt{x}+\sqrt{y}}{2}\le \sqrt{\frac{x+y}{2}}$, which is equivalent to $(\sqrt{x}+\sqrt{y})^2\le 2(x+y)$, which is equivalent to $(\sqrt{x}-\sqrt{y})^2\ge 0$. $\endgroup$ – André Nicolas Mar 24 '15 at 16:06

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 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.


This is no accident actually. In a broader context, the inequality is known as generalized mean inequality, check this wonderful wikipedia page:


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).


  • $\begingroup$ This is an old question which has a well accepted answer. You have contributed nothing new. Please refrain from answering old questions if you are not going to contribute anything new. There are several unanswered questions to choose from $\endgroup$ – Shailesh Jul 19 '16 at 0:57
  • 1
    $\begingroup$ Sorry, new here. I noticed the time after answering it, my bad. However, I do not agree with you that there is nothing new in my answer. The link I posted contains a more general case of this question, and actually a profound result, i.e. the power mean inequality. One might find that pleasing. $\endgroup$ – boyangumn Jul 19 '16 at 4:07
  • $\begingroup$ No problem. ... $\endgroup$ – Shailesh Jul 19 '16 at 4:09

Many such inequalities between the average of a function and the function of the average follow directly from Jensen's inequality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.