A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
Who's up for a challenge? Show that for all positive \( x, y, z \) we have
\[ \left(\frac{x+y}{x+y+z}\right)^{1/2} + \left(\frac{x+z}{x+y+z}\right)^{1/2} + \left(\frac{y+z}{x+y+z}\right)^{1/2} \leq \ 6^{1/2} \]
 2 years ago
Who's up for a challenge? Show that for all positive \( x, y, z \) we have \[ \left(\frac{x+y}{x+y+z}\right)^{1/2} + \left(\frac{x+z}{x+y+z}\right)^{1/2} + \left(\frac{y+z}{x+y+z}\right)^{1/2} \leq \ 6^{1/2} \]

This Question is Closed

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0I can reduce it down to \[(x+y)^{1/2}(y+z)^{1/2} + (x+y)^{1/2}(x + z)^{1/2} + (y+z)^{1/2}(x+z)^{1/2} \le 2(x+y+z)\]

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

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

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

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.5Nicely done. I'll wait a bit and see what other proofs turn up, if any, and then I'll post my solution.

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

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

JamesJ
 2 years ago
Best 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

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

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

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0fool got your wing clipped
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.