## JamesJ 4 years ago Who's up for a challenge? Show that for all positive $$x, y, z$$ we have $\left(\frac{x+y}{x+y+z}\right)^{1/2} + \left(\frac{x+z}{x+y+z}\right)^{1/2} + \left(\frac{y+z}{x+y+z}\right)^{1/2} \leq \ 6^{1/2}$

1. anonymous

I can reduce it down to $(x+y)^{1/2}(y+z)^{1/2} + (x+y)^{1/2}(x + z)^{1/2} + (y+z)^{1/2}(x+z)^{1/2} \le 2(x+y+z)$

2. anonymous

x+y+z$\le$2$1/2$

3. anonymous

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

4. JamesJ

Hint: Use the Cauchy-Schwarz Inequality

5. anonymous

$\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}$

6. JamesJ

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

7. JamesJ

got this ffm?

8. anonymous

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

9. JamesJ

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

10. JamesJ

$\left( \frac{x+y}{x+y+z} \right)^{1/2} + \left( \frac{y+z}{x+y+z} \right)^{1/2} + \left( \frac{z+x}{x+y+z} \right)^{1/2}$ $= 1.\left( \frac{x+y}{x+y+z} \right)^{1/2} + 1.\left( \frac{y+z}{x+y+z} \right)^{1/2} + 1.\left( \frac{z+x}{x+y+z} \right)^{1/2}$ 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

11. anonymous

That's even more compact, well done James.

12. JamesJ

It's a nice problem.

13. anonymous

Indeed it is.

14. anonymous

Jamesj?

15. anonymous

Hello?

16. anonymous

To ping somone use @ feature, for example @JamesJ

17. anonymous