Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Can you prove this trigonometric inequality?

In acute-angled triangle ABC, prove that [CosA/CosB]² + [CosB/CosC]² +[CosC/CosA]² + 8 CosA CosB CosC ≥ 4.

This was originally asked by Annie G without the condition of the triangle being acute-angled and gianlino had shown that the inequality does not hold for all triangles giving an example taking angles of the triangle as 30°, 30° and 120°. He informed me that the inequality is indeed true with the condition that the triangle is acute-angled, but I do not know the proof.

Update:

I have extended time as I am out f station. Shall be back home by 10th and select BA after reading all answers.

6 Answers

Relevance
  • 8 years ago
    Favorite Answer

    For x real x^2 >= 2x - 1 and for y real >0

    (2y - 1)^2 >= 0 ===> 4y^2 - 4y + 1 >= 0 ===> 1/y > 4 - 4y.

    Combining both we get

    [CosA/CosB]^2 >= 2(cos A / cos B) - 1 >= 2 cos(A) (4 - 4 cosB) - 1

    = 8cos A - 8 cos A cos B - 1.

    Adding the three corresponding terms you want

    8 (cos A + cos B + cos C) - 8 (cos A cos B + cos C cos B + cos A cos C) - 3

    + 8 cos A cos B cos C >= 4

    This can be rewritten as 1 >= 8 (1- cos A)(1 - cos B)(1- cos C)

    or 1/64 >= (sin(A/2) sin(B/2) sin(C/2))^2

    This in turn follows from the inequality sin x sin y <= (sin((x+y)/2))^2. This inequality implies that the max of (sin(x) sin(y) sin(z))^2 for x,y,z in [0,pi/2] and x + y + z = pi/2 is attained for x = y = z = pi / 6

    and equals 1/2^6 = 1/64.

  • sri
    Lv 5
    8 years ago

    Nice one. My gut feel was to start with Jenson's inequality,

    A+B+C=π then cosA+cosB+cosC ≤ 3/2

    I went in a couple of seemingly promising ways, but did not reach critical mass (I guess you probably may have tried something on those same lines, Madhukarji).

    Will try again and edit with the solution, hopefully before someone else gets it.

    Whoa, too late.. Looks like gianlino has come up with the solution. Maybe he deserves the BA.

  • 8 years ago

    I think at the first look it needs AM ≥ GM

    LHS = [(cosA/cosB)+(cosB/cosC)+cosC/cosA)]^2 - 6 + 8cosAcosBcosC

    ≥9 - 6 + 8cosAcosBcosC

    But For want of time I will come back to this and solve it.

  • ?
    Lv 4
    5 years ago

    This is a shortcut. Meanwhile i'll try to feel of a more rational manner. Put A = B = C = 60 on all sides of the eq LHS is 1/2*half*1/2 = 1/8 RHS is 1/eight LHS = RHS now put A = B = 60 and C = 90 LHS = 1/4 RHS = 0 LHS > RHS consequently LHS >=RHS

  • How do you think about the answers? You can sign in to vote the answer.
  • 6 years ago

    hii please can anyone solve this:-

    a car A starts from a point P towards another point Q.Another car B starts (also from p) 1 h after the first car and overtakes it after covering 30% of the distance PQ .after that the cars continue and on reaching Q, Car B reverses and meet car A at 23(1/3) of the distance QP.find the time taken by car B to cover the distance PQ(in hours)

  • Anonymous
    8 years ago

    Yeah, it's a tricky one. I couldn't figure it out but I have some links with the proof. This question is the same as question 50 in

    http://adria.inaoep.mx/~diplomados/biblio/GPyT/103...

    The proof there also cites results from

    http://adria.inaoep.mx/~diplomados/biblio/GPyT/103...

Still have questions? Get your answers by asking now.