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Inequality Proof

  1. Apr 14, 2007 #1
    1. The Problem

    For all positive real values of a,b and c such that a²+b²+c²=1, prove that
    a(b+c) < [1/sqrt(2)]

    2. My Attempt...

    There are probably numerous ways to solve this, but i tried this way...

    let a = sin(x)
    b = cos(x)sin(y)
    c = cos(x)cos(y)

    a²+b²+c² = sin²(x)+cos²(x)[sin²(y)+cos²(y)] = 1
    This satisfies the above

    so a(b+c) = sin(x)cos(x)[sin(y)+cos(y)]

    I am sure you can somehow prove that the max value of the give is less than than the value given in the question i.e. 1/sqrt(2)...

    I would appreciate any help... with any other methods.
    Last edited: Apr 14, 2007
  2. jcsd
  3. Apr 14, 2007 #2


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    Write sin(y) + cos(y) in the form A sin(y+p).
  4. Apr 14, 2007 #3
    Sorry I'm quite weak in trigonometry... could you elaborate?
  5. Apr 14, 2007 #4


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    Csin(y+ B)= Csin(y)cos(B)+ Ccos(y)sin(B) = C cos(B)sin(y)+ C sin(B)cos(y)

    You want to find C and B so that C cos(B)= 1 and C sin(B)= 1. Of course, then C2cos2(B)+ C2sin2(B)= C2= 1+ 1= 2. That is, [itex]C= \sqrt{2}[/itex] so that C cos(B)= 1 becomes [itex]\sqrt{2}cos(B)= 1[/itex] and
    [tex]cos(B)= \frac{1}{\sqrt{2}}[/tex]
  6. Apr 14, 2007 #5
    Any other simpler methods?
  7. Apr 14, 2007 #6


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    Your method looks pretty simple.

    The "plug and pray" method would be to find the extreme values of f(a,b,c) = a(b+c) with the constraint g(a,b,c) = a^2+b^2+c^2 =1 using the Lagrange Multiplier method.

    Personally I think your way is neater for this problem.
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