ineq trig4

[Lagrangia]

Prove that for x in (0, pi/2) we have 2cos(x)/(1+cos(x))<sin(x)/x

[liyi]

this ineq is equivalent to
2x < sin x + tan x

since x/2 < tan x/2
it is sufficient to prove that 4 tan x/2 < sin x + tan x
tan x/2 = (sin x)/(1+cos x)
we need only to prove that
4sin x/(1+cos x) < sin x + sin x/cos x
<==>
4cos x < (cos x + 1)^2
which holds obviously.

[liyi]

in fact,
1/(1/(sin x) + 1/(tan x)) = sin x/(1+cos x) = tan x/2 > x/2
and by AM-GM-HM ineq,
(sin x + tan x)/2 > 2/(1/(sin x) + 1/(tan x)) > x
i.e.,
2x < sin x + tan x

[liyi]

and there's a geometric proof for
x < (sin x + tan x)/2

Suppose there's a unit circle O and AB is an arc on it with angle AOB = x.
Let the tagent line passing through A meet the extension of OB at D, and the tangent lines passing A and B intersect at point C. Thus, the area of sector OAB, triangle OAB and triangle OAD are
x/2, sin x/2 and tan x/2
respectively. then we should prove
S_{sector OAB} < (S_{triangle OAB} + S_{triangle OAD})/2

The quadrilateral OACB includes the sector OAB. it is sufficient to prove that
S_{OACB} < (S_{triangle OAB} + S_{triangle OAD})/2
<==>
S_{OACB} - S_{triangle OAB} < S_{triangle OAD} - S_{OACB}
<==>
S_{triangle ACB} < S_{triangle CDB}.

The two triangles, ACB and CDB, have a common vertex B, their edges AC and CD are on the same line, hence they have the same height. To proof the last inequality, we need only to prove that AC<CD. This is correct, because in the right triangle CDB, CD(hypotenuse)>CB.

[liyi]

Remark.
Proving x < (sinx + tanx)/2 is a problem in Hungrian MO 1909.

[Namdung]

We can prove a little bit stronger inequality
2sinx + tgx > 3x for 0 < x < Pi/2.
Proof: Let consider f(x) = 2sinx + tgx - 3x
Then f'(x) = 2cosx + 1/(cosx)^2 - 3 >=0 (by AM-GM 2cosx + 1/(cosx)^2 = cosx + cosx + 1/(cosx)^2 >=3)
So f(x) >= f(0) = 0.

Namdung

[liyi]

well, this one is absolutely ok.
but i think the problems posted here need elementary solutions.

[liyi]