Question

Who's up for a challenge? Show that for all positive \( x, y, z \) we have
\[ \left(\frac{x+y}{x+y+z}\right)^{1/2} + \left(\frac{x+z}{x+y+z}\right)^{1/2} + \left(\frac{y+z}{x+y+z}\right)^{1/2} \leq \ 6^{1/2} \]

Answer 1

I can reduce it down to

\[(x+y)^{1/2}(y+z)^{1/2} + (x+y)^{1/2}(x + z)^{1/2} + (y+z)^{1/2}(x+z)^{1/2} \le 2(x+y+z)\]

Answer 2

x+y+z\[\le\]2\[1/2\]

Answer 3

x=2
y=2
z=2
just one example

Answer 4

Hint: Use the Cauchy-Schwarz Inequality

Answer 5

\[\frac{1}{2}\left(x+y+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{x+y}{x+y+z}}\]\[\frac{1}{2}\left(y+z+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{y+z}{x+y+z}}\]\[\frac{1}{2}\left(z+x+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{z+x}{x+y+z}}\]\[\sum\quad\Rightarrow\quad\frac{1}{2}\left(2(x+y+z)+\frac{3}{x+y+z}\right)\ge\]\[\ge\sqrt{\frac{x+y}{x+y+z}}+\sqrt{\frac{y+z}{x+y+z}}+\sqrt{\frac{z+x}{x+y+z}}\]\[u=x+y+z\quad,\quad f(u)=u+\frac{3}{2u}\quad,\quad f'(u)=1-\frac{3}{2u^2}=0\]\[u=\sqrt{\frac{3}{2}}\quad\Rightarrow\quad f_{\min}=\sqrt{6}\]

Answer 6

Nicely done.

I'll wait a bit and see what other proofs turn up, if any, and then I'll post my solution.

Answer 7

got this ffm?

Answer 8

I think I would have used AM-GM as shown by nikvist.

Answer 9

I'm going to write this out. It's going to take a few minutes ...

Answer 10

\[ \left( \frac{x+y}{x+y+z} \right)^{1/2} + \left( \frac{y+z}{x+y+z} \right)^{1/2} + \left( \frac{z+x}{x+y+z} \right)^{1/2} \]
\[ = 1.\left( \frac{x+y}{x+y+z} \right)^{1/2} + 1.\left( \frac{y+z}{x+y+z} \right)^{1/2} + 1.\left( \frac{z+x}{x+y+z} \right)^{1/2} \]
Now by Cauchy-Schwatz this expression is
\[ \leq (1^2 + 1^2 + 1^2)^{1/2} . \left( \frac{x+y}{x+y+z} + \frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} \right)^{1/2} \]
\[ = 3^{1/2} \left( \frac{2(x+y+z)}{x+y+z} \right)^{1/2} \]
\[ = 6^{1/2} \]
qed

Answer 11

That's even more compact, well done James.

Answer 12

It's a nice problem.

Answer 13

Indeed it is.

Answer 14

Jamesj?

Answer 15

Hello?

Answer 16

To ping somone use @ feature, for example @JamesJ

Answer 17

fool got your wing clipped