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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 \(\frac 32 \)
 one year ago
 one year 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 \(\frac 32 \)
 one year ago
 one year ago

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FoolForMathBest 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.
 one year ago

asnaseerBest 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:\[\sqrt{1\frac{1}{x^2+1}}+\sqrt{1\frac{1}{y^2+1}}+\sqrt{1\frac{1}{z^2+1}}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
which means we need to maximize each of the denominators
 one year ago

asnaseerBest 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
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
but I cannot prove my /gut/ is correct :)
 one year ago

FoolForMathBest 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.
 one year ago

asnaseerBest 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
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
AMGM? CauchySchwarz? Chebyshev? (These are the only ones I know.)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I guess thats a no to both blockcolder and myself :(
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
I like the way asnaseer approaches any problem, always starting from scratch :)
 one year ago

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

ninhi5Best ResponseYou've already chosen the best response.0
\[x \sqrt{y ^{2}+1}\sqrt{z ^{2}+1} = x \sqrt{(y ^{2}+1)(z ^{2}+1})\]
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
\[x \sqrt{y ^{2}+y ^{2}z ^{2}+z ^{2}+y ^{2}z ^{2}}\]
 one year ago

luisrock2008Best ResponseYou've already chosen the best response.0
Your a liar, This is hard
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
Oh, jensen's. Therefore, asnaseer's approach is worth looking at.
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
x(y+z) do the same thing for all of them
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
If you know jensen's this is just there lines.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
never had the pleasure of meeting him :)
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
yeah you guys can just leave me do this alone
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
You were here at 1906 ? :P
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
of course  weren't you?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
19:06 == 6 minutes past 7pm
 one year ago

ninhi5Best 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
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
\[2({\sqrt{3}y}) + 2x(\sqrt{3}  x y)\]
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
\[2\sqrt{3}  2y + 2{\sqrt{3}}x  2x ^{2}  2xy\]
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
differentiate to find max value
 one year ago

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

asnaseerBest 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.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
This is actually concave. f''(x)<0
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I /think/ the sum pf your lambda's equate equate to: a+ b + c = 1
 one year ago

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

asnaseerBest 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
 one year ago

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

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

asnaseerBest ResponseYou've already chosen the best response.1
so now the x', y' and z' correspond to your lambda's blockcolder
 one year ago

asnaseerBest 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.
 one year ago
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