Prove CosA/CosB + CosB/CosC +CosC/CosA + 8 CosA CosB CosC ≥ 4.?

To simplify notations let's use the variables x = cos(A), y = cos(B), z = cos(C). Take the square of the original inequality,

(x/y)^2 + (y/z)^2 + (z/x)^2 + 2*( x/z + z/y + y/x) >= 16 (1- 2 x y z)^2.

Since (x/y)^2 >= 2(x/y) - 1, it is sufficiently to prove

(1): 2( x/y + y/z + z/x + x/z + z/y + y/x) - 3 >= 16 (1- 2 x y z)^2.

This is done in two steps.

STEP 1

Consider an isosceles triangle. Due to symmetry we can take x=y, and z = cos(π-A-B) = - cos(2A) = 1 -2 x^2. Then x/y = 1,

y/z + z/x - 2 = x/(1-2 x^2) + (1-2 x^2)/x - 2 = 4 (x+1)^2 ( x-1/2)^2/( x (1-2 x^2) ),

so that the left side of (1) is equal to

16 (x + 1)^2 (x - 1/2)^2/(x (1 - 2 x^2)) + 9.

Subtracting 9 from the right side we get

16 (1- 2 x y z)^2 - 9 = 16 (x-1/2)^2 (x+1/2)^2 (7 - 8 x^2 + 16 x^4),

and the inequality reduces to

(x - 1/2)^2/(x (1 - 2 x^2)) [ (x + 1)^2 - x (1-2 x^2) (x+1/2)^2 (7 - 8 x^2 + 16 x^4)] >= 0.

Since A=B >= π/4 and x^2 < 1/2, it is sufficient to prove that

(x + 1)^2 - 7 x (1-2 x^2) (x+1/2)^2 >=0.

Expand the second term:

(1-2 x^2) (x +1/2)^2 = -2x^4 - 2x^3 +(1/2) x^2 + x + 1/4 =

[-2x^4 - 2x^3 +(3/2) x^2 + x + 1/4] - x^2.

The derivative of the term in square brackets is

- 8x^3 - 6x^2 + 3x + 1 = (1 - 2 x) (1+x) (1+4 x).

The value -2x^4 - 2x^3 +(3/2) x^2 + x + 1/4 has maximum at x = 1/2 that is equal to 3/4. Then it is sufficient to prove that

(x + 1)^2 - 7 x (3/4 - x^2) = 7 x^3 + x^2 - (13/4) x + 1 >= 0.

Taking derivative we and solving the resulting quadratic equation we see that the minimum is reached at x = ( (4+13*21)^(1/2)-2)/42 and it is approximately equal to 0.285>0.

Thus, inequality (1) holds in isosceles triangles.

STEP 2

The left side of (1) is

2( x/y + y/z + z/x + x/z + z/y + y/x) - 3 =

2( x+y+z) (1/x + 1/y + 1/ z) - 9 =

2( x+y+z) (x y + x z + y z) /(x y z) - 9.

Since A, B, and C are angles of a triangle, they satisfy the equation (take square of sin(A) sin(B) = cos(A) cos(B) + cos (C) and replace sin^2 by 1-cos^2)

2 x y z = 1 - (x^2 + y^2 + z^2), or

2 p = 2 (x y + x z + y z) = (x + y + z)^2 + 2 x y z - 1.

Our inequality reduces to

F(s,q) = s ( s^2 + 2q - 1) - q ( 9 + 16 (1- 2 q)^2 ) =

-25 q + 64 q^2 - 64 q^3 - s + 2 q s + s^3 >= 0.

where s = x + y + z, q = x y z.

Variables s, p, and q have been introduced in such a way that according to Vieta's formulas x, y, z are roots of the cubic polynomial

P(X) = X^3 - s X^2 + p X - q.

This roots are real if the discriminant D of P(X),

D = -27q^2 + 18 s p q + p^2 s^2 - 4 s^3 q - 4 p^3 is non-negative.

Taking p = ( s^2 + 2 q - 1)/2 one can determine the area S in the (s,q)-plane where D>=0. Inside this area

dF/dq = - 25 + 64 q ( 2 - 3 q) + 2 s <= - 25 + 64 *2 q + 2 s.

Taking into account that q<=1/8 and s<=3/2, we have dF/dq < -6. This means that the minimum of F is reached at the boundary of the area, that is composed by the curves q = - (1+s)^2/4, q = (1/2) [-5 +5 s - s^2 +/- √((3-2 s)^3)], and the condition F >= 0 should be first violated there. As is explained in

http://www.emis.de/journals/JIPAM/images/052_07_JI...

(see Figure 1) the boundary of S corresponds to degenerate triangles (with a zero angle), right triangles, and isosceles triangles. In degenerate and right triangles q=0, and F(s,q) = s (s^2 -1) >=0 since s >=1. The isosceles triangles have been checked in STEP 1.

Thus the inequality considered holds everywhere.

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The above estimates can apparently be improved. I think that a stricter inequality

cos(A)/cos(B) + cos(B)/cos(C) + cos(C)/cos(A) + 16 cos(A) cos(B) cos(C) >= 5

is also true.

As in prob 50 vide the adv link provided by cpmb the problem narrows down to proving cosA/cosB+cosB/cosC+cosC/cosA>4(cossqA+cossqB+cossqC).

By am>gm it can be shown in a similar manner as in prob50 that cosA/cosB+cosB/cosC+cosC/cosA >=2(cosA+cosB+cosC).

so it further narrows down to proving 2(cosA+cosB+cosC)>=4(cossqA+cossqB+cossqC) or cosA+cosB+cosC>=2(cossqA+cossqB+cossqC) or 2(cossqA/2+cossqB/2+cossqC/2 - 3/2) >=2(cossqA+cossqB+cossqC). Now the inequality holds for any acute angled triangle where each of cossqA/2-1/2 >=cossqA hence proved.

As in prob 50 vide the adv link offered via skill of cpmb the dilemma narrows each and every of how right down to proving cosA/cosB+cosB/cosC+cosC/cosA>4(cossqA+c... via skill of am>gm this is shown in a matching approach as in prob50 that cosA/cosB+cosB/cosC+cosC/cosA >=2(cosA+cosB+cosC). So it extra narrows each and every of how right down to proving 2(cosA+cosB+cosC)>=4(cossqA+cossqB+co... Or cosA+cosB+cosC>=2(cossqA+cossqB+cossqC) or 2(cossqA/2+cossqB/2+cossqC/2 - 3/2) >=2(cossqA+cossqB+cossqC). Now the inequality holds for any acute angled triangle the area each and every of cossqA/2-a million/2 of >=cossqA for this reason proved.

cos is not negative