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Proving Inequalities

  1. Feb 2, 2005 #1
    Hello all

    How would you prove the following:

    (a) [tex] x + \frac{1}{x} \geq 2, x > 0 [/tex]
    (b) [tex] x + \frac{1}{x} \leq -2, x < 0 [/tex]
    (c) [tex] |x+\frac{1}{x}| \geq 2, x\neq 0. [/tex]

    For all of these inequalities would I simply solve for x, or would I have to use things like the triangle inequality of Schwarz's Inequality?

    Ok I was also given some harder problems with the same concepts:Prove the following inequalities:
    (i) [tex] x^2 + xy + y^2\geq 0 [/tex]
    (ii) [tex] x^{2n} +x^{2n-1}y + x^{2n-2}y^2+ ... + y^{2n} \geq 0 [/tex]
    (iii) [tex] x^4 - 3x^3 + 4x^2 - 3x + 1 \geq 0. [/tex]

    Ok for (i) we can factor [tex]\frac{x^3-y^3}{x-y} [/tex]. If [tex] x > y, x < y [/tex] this expression is positive. For [tex] x = y [/tex] we have [tex] 3x^2 [/tex] which is positive.

    For (ii) we have the same thing except [tex] \frac{x^{2n+1} - y^{2n+1}}{x-y} [/tex].

    For (iii) [tex] x^4 - 3x^3 + 4x^2 - 3x + 1 \geq 0
    \ [/tex][tex] x^4 - 3x^3 + 4x^2 - 3x + 1 = (x-1)^{2}(x^2-x+1) [/tex]
    [tex]x^2 - x + 1 = (x-\frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4} [/tex]

    Would the above problems be similar to these?

    Thanks :smile:
     
    Last edited: Feb 2, 2005
  2. jcsd
  3. Feb 2, 2005 #2

    arildno

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    For your first:
    Differentiate x+1/x.
    You should be able to show that for x>0, x=1 is a minimum.
    Do somewhat analogous for x<0
     
  4. Feb 2, 2005 #3
    well actually you are not supposed to use differentiation as this is the first chapter in the calculus book. so would i just solve for x?

    thanks
     
  5. Feb 2, 2005 #4

    arildno

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    In that case, assume for x>0, [tex]x+\frac{1}{x}<2[/tex]
    Show that this leads to a contradiction.

    Make a similar argument in the case x<0
     
    Last edited: Feb 2, 2005
  6. Feb 2, 2005 #5

    HallsofIvy

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    I confess I don't see how differentiating would help.

    To solve [itex]x+ \frac{1}{x}\ge 2[/itex], as long as you know x> 0, you can multiply the entire inequality by x to get [itex]x^2+ 1\ge 2x[/itex] or [itex] x^2- 2x+ 1\ge 0[/itex] That factors as [itex](x-1)^2\ge 0[/itex] which, since a square is always positive, is true for all x.
     
  7. Feb 2, 2005 #6
    ok thanks a lot :smile:
     
  8. Feb 3, 2005 #7

    arildno

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    Eeh, define the function: [tex]f(x)=x+\frac{1}{x}[/tex]
    Hence, [tex]f'(x)=1-\frac{1}{x^{2}}[/tex]
    which shows that [tex]x=\pm1[/tex] are critical points.
    The 2. derivative test shows that x=1 is a local minimum (with f(1)=2), whereas x=-1 is a local maximum f(-1)=-2.

    Furthermore, by inspecting the signs of f' on either side of, say, x=1, we may readily conclude that x=1 is a "global" minimum for x>0 (that is, f(x)>=2, x>0).
     
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