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If a + b + c = abc, prove that at least one of (a,b,c) is < or = sqrt(3)

  1. Jun 5, 2004 #1

    Here's a question....

    Let a, b, c be three non-zero real numbers such that

    a + b + c = abc

    Prove that at least one of these three numbers (a, b or c) is less than or equal to the square root of 3.

    Can you prove this without trigonometry? The trigonometric solution follows...

    Solution (using Trigonometry)

    Let a = tan(A), b = tan(B), c = tan(C) (for some nonzero angles A, B, C which are real) so that the given constraint becomes

    tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C)

    which can be true iff A + B + C = n*pie (n is an integer)

    If n = 1, then A, B, C are the angles of a triangle (as the constraint is true for angles of a triangle). The result follows by considering cases: of an equilateral triangle where A = B = C = pie/3 radians so that each of a, b and c is equal to sqrt(3); next consider the case of a scalene triangle where A, B and C are all distinct. If A > pie/3, then B+C = pie-A = pie-(qty less than pie/3) and so either B or C is less than pie/3.


  2. jcsd
  3. Jun 5, 2004 #2


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    i) [tex]a=b=c=\sqrt{3}[/tex] is a solution of the equation
    ii) Setting [tex]a=\sqrt{3}+\hat{a}[/tex] and similarly for b and c

    The proof follows readily from this
    (i.e., at least one of the hatted numbers must be non-positive.)
    Last edited: Jun 5, 2004
  4. Jun 6, 2004 #3

    Thanks. Your method is interesting...we get a new constraint on the hatted numbers now...am I right? (I haven't completed the solution using your substitutions..)

  5. Jun 6, 2004 #4

    matt grime

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    proof without using trig or substitution: use the arithmetic mean/geometric mean inequality
  6. Jun 6, 2004 #5


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    Yes, you end up with an equation which can be written like

    [tex] f(\hat{a},\hat{b},\hat{c}) = -g(\hat{a},\hat{b},\hat{c}) [/tex]

    And you'll find that for positive values of [tex] \hat{a},\hat{b},\hat{c}[/tex], the functions f and g must give positive numbers. So you have LHS = positive number and RHS = negative number...a contradiction ! So, one of [tex] \hat{a},\hat{b},\hat{c}[/tex] must not be positive.
    Last edited: Jun 6, 2004
  7. Jun 6, 2004 #6
    Thanks Gokul43201

    I get it now :-D

  8. Jun 6, 2004 #7
    I don't =[
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