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Math Help - some inequalities

  1. #1
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    some inequalities

    started college and havent been doing so good and well i have this test soon and im having trouble with some questions... (i appologize in advance i have to translate from spanish... so if something sounds funny well ask me)

    1) Determine the dimensions of the rectangle with the greatest area which perimeter is 8.

    2) If x, y > 0. prove that:

    2/((1/x)+(1/y)) <= sqrt(xy) <= (x+y)/2

    3) If a, b, c are positive real numbers and different of each other, demonstrate that:

    (a + b + c) / 3 > ((ab + bc + ca)/3)^(1/2) > (abc)^(1/3)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sacwchiri View Post
    started college and havent been doing so good and well i have this test soon and im having trouble with some questions... (i appologize in advance i have to translate from spanish... so if something sounds funny well ask me)

    1) Determine the dimensions of the rectangle with the greatest area which perimeter is 8.
    This is an optimization problem. See the diagram below


    The Calculus way to do this problem.


    we have P = 2x + 2y = 8 ............P is the perimeter
    and we want the area A = xy to be a maximum

    from P, we have
    2x + 2y = 8
    => x = 4 - y ..........solved for x

    => A = xy = (4 - y)y = 4y - y^2

    for max area, we want A' = 0
    => A' = 4 - 2y = 0
    => 4 - 2y = 0
    => y = 2

    but x = 4 - y
    => x = 4 - 2
    => x = 2

    so for our rectangle, x = y = 2, so we want a 2 x 2 rectangle


    The Precalculus way to do this problem

    we have P = 2x + 2y = 8 ............P is the perimeter
    and we want the area A = xy to be a maximum

    from P, we have
    2x + 2y = 8
    => x = 4 - y ..........solved for x

    => A = xy = (4 - y)y = 4y - y^2

    For max A, we want the vertex of the parabola 4y - y^2 = 0

    vertex occurs when y = -b/2a
    => we want y = -4/2(-1) = -4/-2 = 2
    so y = 2

    but x = 4 - y
    => x = 4 - 2
    => x = 2

    so for our rectangle, x = y = 2, so we want a 2 x 2 rectangle

    got it?
    Attached Thumbnails Attached Thumbnails some inequalities-question-1.gif  
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    Quote Originally Posted by sacwchiri View Post
    1) Determine the dimensions of the rectangle with the greatest area which perimeter is 8.
    That rectangle is actually a square.
    2) If x, y > 0. prove that:

    2/((1/x)+(1/y)) <= sqrt(xy) <= (x+y)/2
    There is nothing to prove.
    This is the
    HM-GM-AM
    Inequality.
    3) If a, b, c are positive real numbers and different of each other, demonstrate that:

    (a + b + c) / 3 > ((ab + bc + ca)/3)^(1/2) > (abc)^(1/3)
    By Cauchy-Swartz Inequality (my personal favorite)
    When a,b,c are different.

    a^2+b^2+c^2 > ab+bc+ac

    a^2+b^2+c^2+2ab+2ac+2bc > 3ab+3bc+3ac

    (a^2+2ab+b^2)+2c(a+b)+c^2 > 3ab+3bc+3ac

    (a+b)^2+2(a+b)c+c^2 > 3ab+3bc+3ac

    (a+b+c)^2 > 3ab+3bc+3ac

    (a+b+c)^2/9 > (ab+bc+ac)/3

    Take square roots,

    (a+b+c)/3 > ((ab+bc+ac)/3)^(1/2)
    Q.E.D.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    This is the
    HM-GM-AM
    Inequality.
    what's that? where can i find a proof for it?
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  5. #5
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    axioms and reals

    i got what Jhevon said but for theperfecthcaker what is the hm-gm-am inequality... i should have add im not any good at math...

    and maybe while you are at it explain the Cauchy-Swartz Inequality... cos i dont get that part...

    thanx
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    Heir
    Attached Thumbnails Attached Thumbnails some inequalities-picture13.gif  
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    sooo what you mean is that
    a^2+b^2+c^2 will always be >= ab+bc+ac???

    and then i get kinda lost...
    (sorry for anoying but this is really helping me out now )
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    Quote Originally Posted by sacwchiri View Post
    sooo what you mean is that
    a^2+b^2+c^2 will always be >= ab+bc+ac???

    and then i get kinda lost...
    (sorry for anoying but this is really helping me out now )
    a^2+b^2+c^2 > = ab+bc+ac when a,b,c>0

    Why?

    Explanation.
    -------------

    You can write,
    (a^2+b^2+c^2)(b^2+c^2+a^2) > = (ab+bc+ac)^2
    By Cauchy-Swartzh Inequality.
    Thus,
    (a^2+b^2+c^2)^2 >= (ab+bc+ac)^2
    Take square roots (and note that a,b,c>0)

    a^2+b^2+c^2 >= ab+bc+ac
    Since,
    a != b!=c
    We cannot have equality and hence,
    a^2+b^2+c^2 > ab+bc+ac
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