6

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.

Michael Rozenberg
  • 203,855
Richard
  • 313
  • Does it help to take power two on both sides? – Sigur Jul 31 '13 at 16:33
  • I did but it just gets more complicated. – Richard Jul 31 '13 at 16:33
  • I mean, take powers on your last simplified inequality. So you'll get inequalities involving product of roots and roots of a sum. – Sigur Jul 31 '13 at 16:35
  • 1
    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. – Richard Jul 31 '13 at 16:40
  • 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. – Scaramouche Jul 31 '13 at 16:56

2 Answers2

7

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.

Scaramouche
  • 1,944
  • 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.     – Scaramouche Jul 31 '13 at 17:03
  • 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. – Richard Aug 01 '13 at 16:51
0

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!

Michael Rozenberg
  • 203,855