Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
FoolForMath
Group Title
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 \(\frac 32 \)
 2 years ago
 2 years ago
FoolForMath Group Title
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 \(\frac 32 \)
 2 years ago
 2 years ago

This Question is Closed

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
we 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:\[\sqrt{1\frac{1}{x^2+1}}+\sqrt{1\frac{1}{y^2+1}}+\sqrt{1\frac{1}{z^2+1}}\]
 2 years ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
which 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
 2 years ago

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

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
is 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
 2 years ago

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

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
I like the way asnaseer approaches any problem, always starting from scratch :)
 2 years ago

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

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
\[x \sqrt{y ^{2}+1}\sqrt{z ^{2}+1} = x \sqrt{(y ^{2}+1)(z ^{2}+1})\]
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
\[x \sqrt{y ^{2}+y ^{2}z ^{2}+z ^{2}+y ^{2}z ^{2}}\]
 2 years ago

luisrock2008 Group TitleBest ResponseYou've already chosen the best response.0
Your a liar, This is hard
 2 years ago

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

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
x(y+z) do the same thing for all of them
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
If you know jensen's this is just there lines.
 2 years ago

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

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
yeah you guys can just leave me do this alone
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
You were here at 1906 ? :P
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
of course  weren't you?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
19:06 == 6 minutes past 7pm
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
xy + zy + yx + zy + zx + zy 2zx + 2xy + 2zy 2y(x+z) + 2zx x+z = sqrt 3  y
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Haha :D
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
\[2({\sqrt{3}y}) + 2x(\sqrt{3}  x y)\]
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
\[2\sqrt{3}  2y + 2{\sqrt{3}}x  2x ^{2}  2xy\]
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
differentiate to find max value
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
I'm exhausted
 2 years ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
I /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.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
This is actually concave. f''(x)<0
 2 years ago

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
I was successful in avoiding the Hessian by keeping things only to one variable.
 2 years ago

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

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

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

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
I'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.
 2 years ago
See more questions >>>
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.