anonymous
  • anonymous
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 \(\frac 32 \)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
And yes we don't need to use Lagrange multiplier. ( http://en.wikipedia.org/wiki/Lagrange_multiplier ) otherwise it will be boring.
asnaseer
  • asnaseer
we could write:\[\frac{x}{\sqrt{x^2+1}}=\sqrt{\frac{x^2}{x^2+1}}=\sqrt{\frac{x^2+1-1}{x^2+1}}=\sqrt{1-\frac{1}{x^2+1}}\]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
  • asnaseer
which means we need to maximize each of the denominators

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asnaseer
  • asnaseer
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
asnaseer
  • asnaseer
but I cannot prove my /gut/ is correct :)
anonymous
  • anonymous
The reason why I liked this problem is the fact that it uses an inequality which is very useful in general.
asnaseer
  • asnaseer
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
blockcolder
  • blockcolder
AMGM? Cauchy-Schwarz? Chebyshev? (These are the only ones I know.)
anonymous
  • anonymous
No.
asnaseer
  • asnaseer
I guess thats a no to both blockcolder and myself :(
anonymous
  • anonymous
I like the way asnaseer approaches any problem, always starting from scratch :)
anonymous
  • anonymous
I was talking about Jensen's Inequality ( http://en.wikipedia.org/wiki/Jensen's_inequality )
anonymous
  • anonymous
\[x \sqrt{y ^{2}+1}\sqrt{z ^{2}+1} = x \sqrt{(y ^{2}+1)(z ^{2}+1})\]
anonymous
  • anonymous
\[x \sqrt{y ^{2}+y ^{2}z ^{2}+z ^{2}+y ^{2}z ^{2}}\]
anonymous
  • anonymous
Your a liar, This is hard
blockcolder
  • blockcolder
Oh, jensen's. Therefore, asnaseer's approach is worth looking at.
anonymous
  • anonymous
x(y+z) do the same thing for all of them
anonymous
  • anonymous
If you know jensen's this is just there lines.
asnaseer
  • asnaseer
never had the pleasure of meeting him :)
anonymous
  • anonymous
yeah you guys can just leave me do this alone
anonymous
  • anonymous
You were here at 1906 ? :P
asnaseer
  • asnaseer
of course - weren't you?
asnaseer
  • asnaseer
19:06 == 6 minutes past 7pm
anonymous
  • anonymous
xy + zy + yx + zy + zx + zy 2zx + 2xy + 2zy 2y(x+z) + 2zx x+z = sqrt 3 - y
anonymous
  • anonymous
Haha :D
anonymous
  • anonymous
\[2({\sqrt{3}-y}) + 2x(\sqrt{3} - x -y)\]
anonymous
  • anonymous
\[2\sqrt{3} - 2y + 2{\sqrt{3}}x - 2x ^{2} - 2xy\]
anonymous
  • anonymous
differentiate to find max value
anonymous
  • anonymous
I'm exhausted
blockcolder
  • blockcolder
Jensen's goes like this, right? \[\lambda_1+\lambda_2+\cdots+\lambda_n=1\\ f''(x)>0\\ \therefore f(\lambda_1x_1+\lambda_2x_2+\cdots+\lambda_nx_n)\leq\lambda_1f(x_1)+\lambda_2f(x_2)+\cdots+\lambda_nf(x_n) \]
asnaseer
  • asnaseer
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.
anonymous
  • anonymous
This is actually concave. f''(x)<0
asnaseer
  • asnaseer
I /think/ the sum pf your lambda's equate equate to: a+ b + c = 1
asnaseer
  • asnaseer
*of
anonymous
  • anonymous
I was successful in avoiding the Hessian by keeping things only to one variable.
asnaseer
  • asnaseer
could we divide everything by \(\sqrt{3}\) to get:\[x'+y'+z'=1\]where:\[x'=\frac{x}{\sqrt{3}}\]etc
asnaseer
  • asnaseer
so then we need to maximize:\[\frac{x'}{\sqrt{x'^2+3}}+...\]
asnaseer
  • asnaseer
sorry, that should be - we need to maximize:\[\frac{x'\sqrt{3}}{\sqrt{3x'^2+1}}+...\]
asnaseer
  • asnaseer
so now the x', y' and z' correspond to your lambda's blockcolder
asnaseer
  • asnaseer
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 check-in sometime tomorrow (or should I say later today) to how how this was solved.

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