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JamesJ
Group Title
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
JamesJ Group Title
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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Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
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

Tala Group TitleBest ResponseYou've already chosen the best response.0
x+y+z\[\le\]2\[1/2\]
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
x=2 y=2 z=2 just one example
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
Hint: Use the CauchySchwarz Inequality
 2 years ago

nikvist Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1}{2}\left(x+y+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{x+y}{x+y+z}}\]\[\frac{1}{2}\left(y+z+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{y+z}{x+y+z}}\]\[\frac{1}{2}\left(z+x+\frac{1}{x+y+z}\right)\ge\sqrt{\frac{z+x}{x+y+z}}\]\[\sum\quad\Rightarrow\quad\frac{1}{2}\left(2(x+y+z)+\frac{3}{x+y+z}\right)\ge\]\[\ge\sqrt{\frac{x+y}{x+y+z}}+\sqrt{\frac{y+z}{x+y+z}}+\sqrt{\frac{z+x}{x+y+z}}\]\[u=x+y+z\quad,\quad f(u)=u+\frac{3}{2u}\quad,\quad f'(u)=1\frac{3}{2u^2}=0\]\[u=\sqrt{\frac{3}{2}}\quad\Rightarrow\quad f_{\min}=\sqrt{6}\]
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
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

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
got this ffm?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
I think I would have used AMGM as shown by nikvist.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
I'm going to write this out. It's going to take a few minutes ...
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
\[ \left( \frac{x+y}{x+y+z} \right)^{1/2} + \left( \frac{y+z}{x+y+z} \right)^{1/2} + \left( \frac{z+x}{x+y+z} \right)^{1/2} \] \[ = 1.\left( \frac{x+y}{x+y+z} \right)^{1/2} + 1.\left( \frac{y+z}{x+y+z} \right)^{1/2} + 1.\left( \frac{z+x}{x+y+z} \right)^{1/2} \] Now by CauchySchwatz 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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
That's even more compact, well done James.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.5
It's a nice problem.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Indeed it is.
 2 years ago

MathDude000 Group TitleBest ResponseYou've already chosen the best response.0
Jamesj?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
To ping somone use @ feature, for example @JamesJ
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
fool got your wing clipped
 2 years ago
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