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FoolForMath
 3 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 \(\frac 32 \)
FoolForMath
 3 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

FoolForMath
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1which means we need to maximize each of the denominators

asnaseer
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1but I cannot prove my /gut/ is correct :)

FoolForMath
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0AMGM? CauchySchwarz? Chebyshev? (These are the only ones I know.)

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

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

FoolForMath
 3 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 )

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

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

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

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

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

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

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

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

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

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

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

ninhi5
 3 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

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

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

ninhi5
 3 years ago
Best ResponseYou've already chosen the best response.0differentiate to find max value

blockcolder
 3 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
 3 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.

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

asnaseer
 3 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

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

asnaseer
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1so now the x', y' and z' correspond to your lambda's blockcolder

asnaseer
 3 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.
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