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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 \(\frac 32 \)

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  1. FoolForMath
    • 3 years ago
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    And yes we don't need to use Lagrange multiplier. ( http://en.wikipedia.org/wiki/Lagrange_multiplier ) otherwise it will be boring.

  2. asnaseer
    • 3 years ago
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    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:\[\sqrt{1-\frac{1}{x^2+1}}+\sqrt{1-\frac{1}{y^2+1}}+\sqrt{1-\frac{1}{z^2+1}}\]

  3. asnaseer
    • 3 years ago
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    which means we need to maximize each of the denominators

  4. asnaseer
    • 3 years ago
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    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

  5. asnaseer
    • 3 years ago
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    but I cannot prove my /gut/ is correct :)

  6. FoolForMath
    • 3 years ago
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    The reason why I liked this problem is the fact that it uses an inequality which is very useful in general.

  7. asnaseer
    • 3 years ago
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    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

  8. blockcolder
    • 3 years ago
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    AMGM? Cauchy-Schwarz? Chebyshev? (These are the only ones I know.)

  9. FoolForMath
    • 3 years ago
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    No.

  10. asnaseer
    • 3 years ago
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    I guess thats a no to both blockcolder and myself :(

  11. FoolForMath
    • 3 years ago
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    I like the way asnaseer approaches any problem, always starting from scratch :)

  12. FoolForMath
    • 3 years ago
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    I was talking about Jensen's Inequality ( http://en.wikipedia.org/wiki/Jensen's_inequality )

  13. ninhi5
    • 3 years ago
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    \[x \sqrt{y ^{2}+1}\sqrt{z ^{2}+1} = x \sqrt{(y ^{2}+1)(z ^{2}+1})\]

  14. ninhi5
    • 3 years ago
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    \[x \sqrt{y ^{2}+y ^{2}z ^{2}+z ^{2}+y ^{2}z ^{2}}\]

  15. luisrock2008
    • 3 years ago
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    Your a liar, This is hard

  16. blockcolder
    • 3 years ago
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    Oh, jensen's. Therefore, asnaseer's approach is worth looking at.

  17. ninhi5
    • 3 years ago
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    x(y+z) do the same thing for all of them

  18. FoolForMath
    • 3 years ago
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    If you know jensen's this is just there lines.

  19. asnaseer
    • 3 years ago
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    never had the pleasure of meeting him :)

  20. ninhi5
    • 3 years ago
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    yeah you guys can just leave me do this alone

  21. FoolForMath
    • 3 years ago
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    You were here at 1906 ? :P

  22. asnaseer
    • 3 years ago
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    of course - weren't you?

  23. asnaseer
    • 3 years ago
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    19:06 == 6 minutes past 7pm

  24. ninhi5
    • 3 years ago
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    xy + zy + yx + zy + zx + zy 2zx + 2xy + 2zy 2y(x+z) + 2zx x+z = sqrt 3 - y

  25. FoolForMath
    • 3 years ago
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    Haha :D

  26. ninhi5
    • 3 years ago
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    \[2({\sqrt{3}-y}) + 2x(\sqrt{3} - x -y)\]

  27. ninhi5
    • 3 years ago
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    \[2\sqrt{3} - 2y + 2{\sqrt{3}}x - 2x ^{2} - 2xy\]

  28. ninhi5
    • 3 years ago
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    differentiate to find max value

  29. ninhi5
    • 3 years ago
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    I'm exhausted

  30. blockcolder
    • 3 years ago
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    Jensen's goes like this, right? \[\lambda_1+\lambda_2+\cdots+\lambda_n=1\\ f''(x)>0\\ \therefore f(\lambda_1x_1+\lambda_2x_2+\cdots+\lambda_nx_n)\leq\lambda_1f(x_1)+\lambda_2f(x_2)+\cdots+\lambda_nf(x_n) \]

  31. asnaseer
    • 3 years ago
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    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.

  32. FoolForMath
    • 3 years ago
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    This is actually concave. f''(x)<0

  33. asnaseer
    • 3 years ago
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    I /think/ the sum pf your lambda's equate equate to: a+ b + c = 1

  34. asnaseer
    • 3 years ago
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    *of

  35. FoolForMath
    • 3 years ago
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    I was successful in avoiding the Hessian by keeping things only to one variable.

  36. asnaseer
    • 3 years ago
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    could we divide everything by \(\sqrt{3}\) to get:\[x'+y'+z'=1\]where:\[x'=\frac{x}{\sqrt{3}}\]etc

  37. asnaseer
    • 3 years ago
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    so then we need to maximize:\[\frac{x'}{\sqrt{x'^2+3}}+...\]

  38. asnaseer
    • 3 years ago
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    sorry, that should be - we need to maximize:\[\frac{x'\sqrt{3}}{\sqrt{3x'^2+1}}+...\]

  39. asnaseer
    • 3 years ago
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    so now the x', y' and z' correspond to your lambda's blockcolder

  40. asnaseer
    • 3 years ago
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    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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