JamesJ
  • JamesJ
Who's up for a challenge? Show that for all positive \( x, y, z \) we have \[ \left(\frac{x+y}{x+y+z}\right)^{1/2} + \left(\frac{x+z}{x+y+z}\right)^{1/2} + \left(\frac{y+z}{x+y+z}\right)^{1/2} \leq \ 6^{1/2} \]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I can reduce it down to \[(x+y)^{1/2}(y+z)^{1/2} + (x+y)^{1/2}(x + z)^{1/2} + (y+z)^{1/2}(x+z)^{1/2} \le 2(x+y+z)\]
anonymous
  • anonymous
x+y+z\[\le\]2\[1/2\]
anonymous
  • anonymous
x=2 y=2 z=2 just one example

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JamesJ
  • JamesJ
Hint: Use the Cauchy-Schwarz Inequality
nikvist
  • nikvist
\[\frac{1}{2}\left(x+y+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{x+y}{x+y+z}}\]\[\frac{1}{2}\left(y+z+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{y+z}{x+y+z}}\]\[\frac{1}{2}\left(z+x+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{z+x}{x+y+z}}\]\[\sum\quad\Rightarrow\quad\frac{1}{2}\left(2(x+y+z)+\frac{3}{x+y+z}\right)\ge\]\[\ge\sqrt{\frac{x+y}{x+y+z}}+\sqrt{\frac{y+z}{x+y+z}}+\sqrt{\frac{z+x}{x+y+z}}\]\[u=x+y+z\quad,\quad f(u)=u+\frac{3}{2u}\quad,\quad f'(u)=1-\frac{3}{2u^2}=0\]\[u=\sqrt{\frac{3}{2}}\quad\Rightarrow\quad f_{\min}=\sqrt{6}\]
JamesJ
  • JamesJ
Nicely done. I'll wait a bit and see what other proofs turn up, if any, and then I'll post my solution.
JamesJ
  • JamesJ
got this ffm?
anonymous
  • anonymous
I think I would have used AM-GM as shown by nikvist.
JamesJ
  • JamesJ
I'm going to write this out. It's going to take a few minutes ...
JamesJ
  • 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
anonymous
  • anonymous
That's even more compact, well done James.
JamesJ
  • JamesJ
It's a nice problem.
anonymous
  • anonymous
Indeed it is.
anonymous
  • anonymous
Jamesj?
anonymous
  • anonymous
Hello?
anonymous
  • anonymous
To ping somone use @ feature, for example @JamesJ
anonymous
  • anonymous
fool got your wing clipped

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