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Finding the minimum value

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Let x, y, z be positive real numbers such that ##x+y+z=1##. Determine the minimum value of
    [tex]\frac{1}{x}+\frac{4}{y}+\frac{9}{z}[/tex]


    2. Relevant equations
    Cauchy-Schwarz inequality maybe?


    3. The attempt at a solution
    Applying the Cauchy Schwarz inequality,
    [tex]\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)\leq (1^2+4^2+9^2)\left( \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)[/tex]
    I don't see how can I solve this.
    [tex]\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}[/tex]
    ##(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)## and ##x+y+z=1##
    [tex]\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{(xy+yz+zx)^2-2xyz}{x^2y^2z^2}[/tex]
    I don't think the above simplification is useful here.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Jun 9, 2013 #2

    Ray Vickson

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    This problem is easily tackled using calculus. In particular, you can use either (1) the Lagrange multiplier method (the easiest); or (2) solve for z (say) in terms of x and y from the constraint equation, then substitute that expression in place of z in the function you want to minimize---giving you two-variable unconstrained minimization problem.
     
  4. Jun 9, 2013 #3
    I have never done calculus questions involving three variables. Is it not possible to do this without calculus?
     
  5. Jun 9, 2013 #4

    I like Serena

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    You can do it using the AM-GM inequality ##\frac{a+b}{2} = \sqrt{ab}## with equality only if ##a=b##.

    \begin{aligned}\frac 1 x + \frac 4 y + \frac 9 z &= \left(\frac 1 x + \frac 4 y + \frac 9 z\right)(x+y+z) \\
    &= (1+4+9) + ... \\
    &\ge (1+4+9) + ... \\
    &= (1+4+9) + 2 \sqrt{1 \cdot 4} + 2 \sqrt{1 \cdot 9} + 2 \sqrt{4 \cdot 9}
    \end{aligned}

    Can you fill in the dots? :wink:
     
  6. Jun 9, 2013 #5
    Hi ILS!!!! :smile:

    [tex]\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)(x+y+z)=14+4\left(\frac{x+z}{y}\right)+9\left(\frac{x+y}{z}\right)+\left(\frac{y+z}{x}\right)[/tex]

    I really can't figure out how you brought those surds in the end. I tried applying AM-GM to the expanded expression but that led to me nowhere. :(
     
  7. Jun 9, 2013 #6

    I like Serena

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    Heya. :smile:

    Try it for instance with ##\frac y x + 4\frac x y##.
     
  8. Jun 9, 2013 #7
    Ah, I think I got it.
    [tex]\frac{y}{x}+4\frac{x}{y}\geq 2\sqrt{1\cdot 4}[/tex]
    [tex]\frac{z}{x}+9\frac{x}{z} \geq 2\sqrt{1\cdot 9}[/tex]
    [tex]9\frac{y}{z}+4\frac{z}{y} \geq 2\sqrt{4\cdot 9}[/tex]
    I think I can add the inequalities as ##x,y## and ##z## are positive real numbers.
    Hence,
    [tex]\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\geq 36[/tex]

    Thanks ILS!!! :smile:

    But how can I do it with Cauchy-Schwarz? I found this problem in a trigonometry book and the author wrote that the problem can be done in one-step using Cauchy-Schwarz. As this is a trigonometry book, the author used trigonometric substitutions to solve the problem but I am curious as to how can I do it using Cauchy-Schwarz. And anyways, I always wanted to learn this inequality because a few questions were asked in my tests and I rarely (or once :tongue2:) had success with them.

    Thanks!
     
  9. Jun 9, 2013 #8

    I like Serena

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    Good! :wink:

    Well, your expression is on the wrong side of the Cauchy-Schwarz inequality (and you also left out a square :eek:).

    Cauchy-Schwarz is:
    $$|(a \cdot b)|^2 \le (a \cdot a) (b \cdot b)$$
    For the right side, you want something like
    $$\left(\frac 1x + \frac 4y + \frac 9z\right)(x+y+z)$$
    Which a and b would fit?
     
  10. Jun 9, 2013 #9
    [strike]Why is it wrong? I think I only forgot to put a 2 in the exponent on LHS of inequality.[/strike] :confused:

    EDIT: Got it. Thanks!
    If I take a as ##(\sqrt{x},\sqrt{y},\sqrt{z})## and b as ##(\frac{1}{\sqrt{x}},\frac{2}{\sqrt{y}},\frac{3}{\sqrt{z}})##, it gives the right answer.

    Thank you again ILS!
     
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