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Hard inequalities problem

  1. Oct 28, 2006 #1
    i need to prove the following:
    1)let a,b,c be acute angles, if tg(a)tg(b)tg(c)=1 then sin(a)sin(b)sin(c)<=1/2sqrt2
    2) prove that for every x,y cos(x^2)+cos(y^2)-cos(xy)<3
    for the second question i tried to use the fact that (x^2+y^2)/2>=xy and the fact that on some intervals the function cos is decreasing, actually what i need to prove is that -cos(xy)<1, cause cosx^2+cosy^2<=2, so i also tried to show that cos(xy) cannot be equal -1, but didnt get much with that.

    for the first i used the cosine law and sine law, but also didnt get far, i need to show that (sin^2a(sin^2b+sin^2c)+sin^2b(sin^2a+sin^2c)+sin^2c(sin^2a+sin^2b)<=sin(a)sin(b)sin(c), but i didnt succeded in it.

    p.s
    if someone wonders why im posting questions of these type is because im reviewing a little bit inequalities.
     
  2. jcsd
  3. Oct 28, 2006 #2

    matt grime

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    tg(a) is what?
     
  4. Oct 28, 2006 #3
    a shortcut for tangans of a, or if you will sina/cosa. (-:
     
  5. Oct 28, 2006 #4

    matt grime

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    It's tan, a well known abbreviation already. Is this some new convention with which I'm completely unfamiliar?
     
  6. Oct 28, 2006 #5
    as far as im aware you can use both tan and tg.
    anyway, do you have any hints about my questions?
     
  7. Oct 28, 2006 #6

    Office_Shredder

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    For the second one, the maximum value cos can attain is 1, and the minimum is -1. So you simply have to show that cos(x2)=cos(y2)=1 and cos(xy)=-1 are not simultaneously possible, because that's the only way to make that equal three (you can't break three with the sum of three cos terms). So for cos to equal 1,[tex]x^2=2\pi n[/tex] and same for y2 Try to analyze what xy can be given that (and noting xy must be a multiple of [tex](2n+1) \pi[/tex] to make your inequality untrue)
     
    Last edited: Oct 28, 2006
  8. Oct 28, 2006 #7

    arildno

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    Have you tried using a Lagrange multiplier approach on the first one?
     
  9. Oct 28, 2006 #8

    matt grime

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    I thought of that too, but it seems a bit too complicated, if you ask me. Here are two proofs of the first one without any analysis/calculus.

    The condition essentially states that sin(a)sin(b)sin(c)=cos(a)cos(b)cos(c). Or that the function is symmetric about (pi/4,pi/4,pi/4) in each variable. Thus that is either going to be the maximum or minimum, a check tells us it is not a minimum.

    If that's too hand wavy, how about this one? Fix a. Clearly for any a the maximum we can attain is when b=c. This applies for any a, in particular the a where the maximum occurs. Thus the maximum occurs at (a,b,b). But this is completely symmetric in the arguments, so a=b=c is the maximum. The value of the maximum is then gotten from the condition which reduces to tan(a)^3=1, or a=pi/4.

    I can't be bothered to make those rigorous - which could be done with an argument involving lagrange multipliers, for example.
     
  10. Oct 28, 2006 #9

    Office_Shredder

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    matt, I think your first argument is less hand wavy than the second, which actually assumes off the bat that there DOES exist a maximum.
     
  11. Oct 28, 2006 #10

    matt grime

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    But there is obviously a maximum by inspection.
     
  12. Oct 29, 2006 #11

    arildno

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    As it happens, using Lagrange multipliers turns out to be very simple.
    It is, however, at least as important to understand how the symmetry arguments used by matt are powerful, and ought to be convincing on their own.

    Here's how we might set up the problem using L.M's:
    Define:
    [tex]f(a,b,c)=\sin(a)\sin(b)\sin(c), g(a,b,c)=f(a,b,c)-\cos(a)\cos(b)\cos(c)[/tex]
    Thus, we are to solve the problem:
    [tex]\nabla{f}=\lambda\nabla{g},g=0[/tex]
    Considering the first equation, we have:
    [tex]\cos(a)\sin(b)\sin(c)=\lambda(\cos(a)\sin(b)\sin(c)+\sin(a)\cos(b)\cos(c))[/tex]
    Rearranging a bit, we may write this equation as:
    [tex]\tan(a)=\frac{1-\lambda}{\lambda}\tan(b)\tan(c)[/tex]
    Similar expressions hold, of course, for tan(b) and tan(c), and multiplying these three equations and utilizing g=0, yields:
    [tex](\frac{1-\lambda}{\lambda})^{3}=1\to\lambda=\frac{1}{2}[/tex]
    OP can finish the exercise on his own.
     
    Last edited: Oct 29, 2006
  13. Oct 29, 2006 #12
    the problem is i can't use lagrange multipliers.

    here's my appraoch:
    we have three acute angles, we can use the cosine law and sine law,
    after im using both of these theorems, i get to this inequality:
    sin(a)sin(b)sin(c)<=1/8[sin^2a/sin^2c+sin^2a/sin^2b+sin^2b/sin^2a+sin^2b/sin^2c+sin^2c/sin^2b+sin^2c/sin^2a]
    p.s
    matt, i feel that both your arguments are hand wavy.
    i understand that sin(a)sin(b)sin(c)=cos(a)cos(b)cos(c) but how do you show that the maximum is achieved when a=b=c=pi/4?
     
  14. Oct 29, 2006 #13
    office shredder, if cos(xy)=-1 cos(x^2)=cos(y^2)=1 then we have
    x^2=2npi y^2=2kpi so xy=2(pi)sqrt(nk) but according to the first equation xy=pi+2cpi so we have 2(pi)sqrt(nk)=pi+2cpi
    4nk=1+2c which is a contradiction cause the rhs is an odd number, and the lhs is even.
    ok, thanks.
     
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