A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Just another super easy problem,
Let \(x, y, z \in \mathbb{R}^+\) such that \(x +y +z =\sqrt{3} \) .
Prove that the maximum value of \(\large \frac x{\sqrt{x^2+1}} +\frac y{\sqrt{y^2+1}} +\frac z{\sqrt{z^2+1}}\) is \(\frac 32 \)
anonymous
 4 years ago
Just another super easy problem, Let \(x, y, z \in \mathbb{R}^+\) such that \(x +y +z =\sqrt{3} \) . Prove that the maximum value of \(\large \frac x{\sqrt{x^2+1}} +\frac y{\sqrt{y^2+1}} +\frac z{\sqrt{z^2+1}}\) is \(\frac 32 \)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And yes we don't need to use Lagrange multiplier. ( http://en.wikipedia.org/wiki/Lagrange_multiplier ) otherwise it will be boring.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1we could write:\[\frac{x}{\sqrt{x^2+1}}=\sqrt{\frac{x^2}{x^2+1}}=\sqrt{\frac{x^2+11}{x^2+1}}=\sqrt{1\frac{1}{x^2+1}}\]so then we want to maximize:\[\sqrt{1\frac{1}{x^2+1}}+\sqrt{1\frac{1}{y^2+1}}+\sqrt{1\frac{1}{z^2+1}}\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1which means we need to maximize each of the denominators

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1which gives me the /gut/ feeling that x=y=z is the solution since we have a constraint on the sum of x, y and z

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1but I cannot prove my /gut/ is correct :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The reason why I liked this problem is the fact that it uses an inequality which is very useful in general.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1is it worth pursuing: let x = a\(\sqrt{3}\) let y = b\(\sqrt{3}\) let z = c\(\sqrt{3}\) so we have: a + b + c = 1

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0AMGM? CauchySchwarz? Chebyshev? (These are the only ones I know.)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1I guess thats a no to both blockcolder and myself :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I like the way asnaseer approaches any problem, always starting from scratch :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was talking about Jensen's Inequality ( http://en.wikipedia.org/wiki/Jensen's_inequality )

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x \sqrt{y ^{2}+1}\sqrt{z ^{2}+1} = x \sqrt{(y ^{2}+1)(z ^{2}+1})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x \sqrt{y ^{2}+y ^{2}z ^{2}+z ^{2}+y ^{2}z ^{2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Your a liar, This is hard

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, jensen's. Therefore, asnaseer's approach is worth looking at.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x(y+z) do the same thing for all of them

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you know jensen's this is just there lines.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1never had the pleasure of meeting him :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah you guys can just leave me do this alone

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You were here at 1906 ? :P

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1of course  weren't you?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.119:06 == 6 minutes past 7pm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0xy + zy + yx + zy + zx + zy 2zx + 2xy + 2zy 2y(x+z) + 2zx x+z = sqrt 3  y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2({\sqrt{3}y}) + 2x(\sqrt{3}  x y)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2\sqrt{3}  2y + 2{\sqrt{3}}x  2x ^{2}  2xy\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0differentiate to find max value

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0Jensen's goes like this, right? \[\lambda_1+\lambda_2+\cdots+\lambda_n=1\\ f''(x)>0\\ \therefore f(\lambda_1x_1+\lambda_2x_2+\cdots+\lambda_nx_n)\leq\lambda_1f(x_1)+\lambda_2f(x_2)+\cdots+\lambda_nf(x_n) \]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1I /think/ you are correct blockcolder from what I've just read about this method, but I don't understand it well enough to apply it here.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is actually concave. f''(x)<0

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1I /think/ the sum pf your lambda's equate equate to: a+ b + c = 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was successful in avoiding the Hessian by keeping things only to one variable.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1could we divide everything by \(\sqrt{3}\) to get:\[x'+y'+z'=1\]where:\[x'=\frac{x}{\sqrt{3}}\]etc

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1so then we need to maximize:\[\frac{x'}{\sqrt{x'^2+3}}+...\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1sorry, that should be  we need to maximize:\[\frac{x'\sqrt{3}}{\sqrt{3x'^2+1}}+...\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1so now the x', y' and z' correspond to your lambda's blockcolder

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1I'm probably way off the mark as its very late here and I need to get some sleep. good like with this guys. I'll checkin sometime tomorrow (or should I say later today) to how how this was solved.
Ask your own question
Sign UpFind 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.