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    Handout 435 535

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    AsriGyu Cullenzious

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    Handout 435 535

    Uploaded by

    AsriGyu Cullenzious
    23 views3 pages

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    Handout-435-535

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    23 views3 pages

    Handout 435 535

    Uploaded by

    AsriGyu Cullenzious

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
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    of 3

    THE AM-GM INEQUALITY

    MATH CIRCLE (ADVANCED) 11/11/2012

    The arithmetic mean (AM ) and geometric mean (GM ) of the numbers a1 , a2 , . . . , an ≥ 0
    are given by
    a1 + · · · + an √
    AM = and GM = n a1 · · · an .
    n
    0) a) Calculate the AM and GM of the following sets of numbers:
    i) 1, 2, 3, 6
    ii) 0, 4, 8, 20
    iii) 1, 3, 4, 7, 8
    iv) 4, 4, 4, 4
    b) What do you notice about AM compared to GM ?
    See ↓
    The AM-GM inequality states that if a1 , a2 , . . . , an ≥ 0, then

    a1 + · · · + an √
    ≥ n a1 · · · an with equality if and only if a1 = a2 = · · · = an .
    n
    2) a) Write out the AM-GM inequality for the numbers a, b.
    b) Prove your statement in a).

    a+b √
    ≥ ab ⇔ (a + b)2 ≥ 4ab ⇔ a2 + 2ab + b2 ≥ 4ab ⇔ a2 − 2ab + b2 ≥ 0 ⇔ (a − b)2 ≥ 0
    2
    3) Prove the following:
    x2 + y 2
    a) ≥ xy for any x, y.
    2
    Use AM-GM inequality with a = x2 , b = y 2 .
    b) 2(x2 + y 2 ) ≥ (x + y)2 for any x, y.
    Use a) and the fact that 2(x2 + y 2 ) ≥ (x + y)2 ⇔ x2 + y 2 ≥ 2xy.
    1 1 4
    c) + ≥ for x, y ≥ 0.
    x y x+y
    Hint: Find a common denominator.
    4) Prove that
    a) x2 + y 2 + z 2 ≥ xy + yz + zx for any x, y, z.
    c
    Copyright 2008-2012 Olga Radko/Los Angeles Math Circle/UCLA Department of Mathematics .
    1
    LAMC handout 2
    Hint: Use 3a) and add up three inequalities.
    b) (a + b)(a + c)(b + c) ≥ 8abc
    Hint: Use the AM-GM inequality three times.
    c) x2 + y 2 + 1 ≥ xy + x + y if x, y ≥ 1, (for a challenge prove it for any x, y!).
    Using 3a) we have x2 + y 2 + 1 ≥ 2xy + 1 so it is enough to prove 2xy + 1 ≥ xy + x + y if
    x, y ≥ 1. We have 2xy + 1 ≥ xy + x + y ⇔ xy − x − y + 1 ≥ 0 ⇔ (x − 1)(y − 1) ≥ 0.
    In general, x2 + y 2 + 1 − xy − x − y = (x − y)2 + (x − 1)2 + (y − 1)2 /2.
    d) x4 + y 4 + z 4 ≥ xyz(x + y + z) for any x, y, z.
    Hint: Use 4a) twice.
    5) The sum of two non-negative numbers is 10. What is the maximum and minimum
    value of the sum of their squares?
    The sum of the squares is x2 + (10 − x)2 = 2x2 − 20x + 100 = 2[(x − 5)2 + 20]. The
    minimum occurs at x = 5 while the maximum occurs at x = 0 or x = 10.
    6) Prove the AM-GM inequality for n = 4.
    Using 2a) twice we have:

    1 √ √  √ √ √
      q
    a+b+c+d 1 a+b c+d 4
    = + ≥ ab + cd ≥ ab cd = abcd
    4 2 2 2 2
    7) Prove that: (Hint: You may use the AM-GM inequality for any n.)
    a) a4 + b4 + 8 ≥ 8ab for any a, b.
    Hint: Apply AM-GM to x = a4 , y = b4 , z = 4, w = 4.
    1 1 1 1
    b) (a + b + c + d)( + + + ) ≥ 16 for a, b, c, d ≥ 0.
    a b c d
    Hint: What does AM-GM tell use about the two terms of the left-hand side?
    c) 3x3 − 6x2 + 4 ≥ 0 for x ≥ 0.
    Hint: Apply AM-GM to a = 2x3 , b = x3 , c = 4.
    a b c
    d) + + ≥ 3 for a, b, c ≥ 0.
    b c a
    Hint: Simply use AM-GM.
    8)
    a) Prove the AM-GM inequality for n = 3.
    Suppose
    √ a, b, c are our three numbers. Applying AM-GM to x = a, y = b, z = c, w =
    3
    abc we have

    x+y+z+w √ √4 a+b+c √ 3
    ≥ 4 xyzw = w4 = w ⇒ x + y + z ≥ 3w ⇒ ≥ abc.
    4 3
    b)* Prove by induction that the AM-GM inequality holds for n a power of 2.
    Hint: Use the same trick as in 6) for the induction step.
    2
    LAMC handout 3

    c)* Prove the AM-GM inequality for all n.


    Hint: Use the same trick as in 8a).
    Some problems are taken from:
    • D. Fomin, S. Genkin, I. Itenberg “Mathematical Circles (Russian Experience)”

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