6
$\begingroup$

I did substitute $a=x+y, b=x+z, c=y+z$ and I arrived at $\sqrt{2x} + \sqrt{2y} + \sqrt{2z} \le \sqrt{x+y} + \sqrt{x+z} + \sqrt{y+z}$. However, after this, I tried various methods like AM-GM and Cauchy-Schwarz inequality for hours and I still can't prove it. Can someone help please? Thanks.

$\endgroup$
5
  • $\begingroup$ Does it help to take power two on both sides? $\endgroup$
    – Sigur
    Jul 31, 2013 at 16:33
  • $\begingroup$ I did but it just gets more complicated. $\endgroup$
    – Richard
    Jul 31, 2013 at 16:33
  • $\begingroup$ I mean, take powers on your last simplified inequality. So you'll get inequalities involving product of roots and roots of a sum. $\endgroup$
    – Sigur
    Jul 31, 2013 at 16:35
  • 1
    $\begingroup$ You mean I should square $\sqrt{2x}+\sqrt{2y}+\sqrt{2z} \le \sqrt{x+y}+\sqrt{x+z}+\sqrt{y+z}$ ? That is what I meant when I said it got more complicated. $\endgroup$
    – Richard
    Jul 31, 2013 at 16:40
  • $\begingroup$ I think I can do it by assuming $x+y+z =1$ and then using "fudging" on the resulting inequality, but I'm also pretty sure there's a "classical" method as well. $\endgroup$ Jul 31, 2013 at 16:56

2 Answers 2

7
$\begingroup$

Since $\sqrt{x}$ is concave down, Jensen's inequality tells us that
$ \dfrac 12 ( \sqrt{2x} + \sqrt{2y}) \leq \sqrt{ \dfrac{ 2x + 2y } 2 } = \sqrt{x+y}$.
Summing cyclically gives the desired result.

$\endgroup$
2
  • $\begingroup$ The motivation on using Jensen's is as follows: when you look at $ \sum \sqrt{2x} \leq \sum \sqrt{x+y}$, it's easy to see that we have equality if the square roots weren't there. It's also worth plotting $2x, 2y, 2z, x+y, x+z, y+z$ to note that they are "about the same size", but $2x, 2y, 2z$ are further away from the average $2/3 (x+y+z)$. So now we have a function evaluated at 3 "outer" points which we want to prove is less than when evaluated at 3 "inner" points, which sounds exactly like Jensen. $\endgroup$ Jul 31, 2013 at 17:03
  • $\begingroup$ In other words, if we square $\frac{1}{2}(\sqrt{2x} + \sqrt{2y}) \le \sqrt{x+y}$, then $ x+y \ge \frac{1}{4}(2x+2y+4(\sqrt{xy}))$ . This leads to $x+y-2\sqrt{xy} \ge 0$ which is true. Sorry, I still have a bit of trouble with Jensen's inequality but thanks for answering. $\endgroup$
    – Richard
    Aug 1, 2013 at 16:51
0
$\begingroup$

Let $a\geq b\geq c$ and $f(x)=\sqrt{x}$.

Hence, $a+b-c\geq a+c-b\geq b+c-a$, $(a+b-c,a+c-b,b+c-a)\succ(a,b,c)$ and since $f$ is a concave function, the starting inequality it's just Karamata.

Done!

$\endgroup$

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.