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JamesJ

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

  • 2 years ago
  • 2 years ago

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  1. Ishaan94
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    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 years ago
  2. Tala
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    x+y+z\[\le\]2\[1/2\]

    • 2 years ago
  3. cinar
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    x=2 y=2 z=2 just one example

    • 2 years ago
  4. JamesJ
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    Hint: Use the Cauchy-Schwarz Inequality

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

    • 2 years ago
  6. JamesJ
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    Nicely done. I'll wait a bit and see what other proofs turn up, if any, and then I'll post my solution.

    • 2 years ago
  7. JamesJ
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    got this ffm?

    • 2 years ago
  8. FoolForMath
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    I think I would have used AM-GM as shown by nikvist.

    • 2 years ago
  9. JamesJ
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    I'm going to write this out. It's going to take a few minutes ...

    • 2 years ago
  10. JamesJ
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    \[ \left( \frac{x+y}{x+y+z} \right)^{1/2} + \left( \frac{y+z}{x+y+z} \right)^{1/2} + \left( \frac{z+x}{x+y+z} \right)^{1/2} \] \[ = 1.\left( \frac{x+y}{x+y+z} \right)^{1/2} + 1.\left( \frac{y+z}{x+y+z} \right)^{1/2} + 1.\left( \frac{z+x}{x+y+z} \right)^{1/2} \] 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

    • 2 years ago
  11. FoolForMath
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    That's even more compact, well done James.

    • 2 years ago
  12. JamesJ
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    It's a nice problem.

    • 2 years ago
  13. FoolForMath
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    Indeed it is.

    • 2 years ago
  14. MathDude000
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    Jamesj?

    • 2 years ago
  15. MathDude000
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    Hello?

    • 2 years ago
  16. FoolForMath
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    To ping somone use @ feature, for example @JamesJ

    • 2 years ago
  17. Libniz
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    fool got your wing clipped

    • one year ago
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