be real numbers . Prove that
![$$-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}\leq \frac{2+\sqrt[3] 2+\sqrt[3] {4}}{12} $$](http://latex.artofproblemsolving.com/9/9/9/9997397eb38cd4bceec2ff8c85951c78e2797d0f.png)
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
and 
.
Y by Amir Hossein, dave_mathsguru, 799786, itslumi, Adventure10, Mango247, fuzzball2023, and 1 other user
Let be given a triangle 
and its internal angle bisector 
. The line intersects the circumcircle 
of triangle at 
and 
. Circle 
with diameter 
cuts again at 
. Prove that 
is the symmedian line of triangle .
and its internal angle bisector 
. The line intersects the circumcircle 
of triangle at 
and 
. Circle 
with diameter 
cuts again at 
. Prove that 
is the symmedian line of triangle .
be midpoint of 
.
,
,
.
then:
,
,
,
(a.a), 
,
.
(thanks to 
,
(Combine 
)
is a harmonic quadrilateral. Therefore, 
,
is the circumcenter of 
.
the midpoint of 
.
is cyclic (let 
its circumcenter) . Then 
are collinear . So if the circumcircle of 
, then 
. So by polars 
is tangent to 
is the 
meets the circumcircle of 
, then 
. But also because 
we have that 
is the incenter of 
so 
and because 
we also have 
. This shows that 
is cyclic and so 
and we are done .
cuts 
at 
are incenter and B-excenter of 
is a circle with center 
The cross ratio 
is harmonic and 
is radical axis of 
Pairwise radical axes 
of 
meet at their radical center 
and 
is cyclic. Its circumcircle is the B-Apollonius circle of 
at 
. Denote the following points : the middlepoint ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
;
for which 
; the point 
; the 
-
,
; the point 
. Then 
, i.e. the quadrilateral 
is cyclically.
- the tangent in 
to 
. Observe that 
and 
.
and the line 
is the 
.
is harmonically. Since 
results 
.
(a.a.) . Thus, 
, i.e. 
.
,
, i.e. 
.
obtain 
and in consequence, 
. Since 
obtain 
, i.e. the quadrilateral 
.
.
the midpoint of segment 
meets 
,we have 
are collinear.
is cyclic and 
is cylic ,we have :
.
.
.
is harmonically,so 
are collinear.
and denote by 
the intersection of two tangent to 
,respectively.
lie on the polar of 
wrt 
,we obtain 
are collinear.
(where 
at a point 
such that 
.![\[ \angle{MBC} = \angle{MTA} = \angle{ABF}.
\]](http://latex.artofproblemsolving.com/2/3/9/239532c7002a5d2f2d9ff6e8e73e2630d0ff7ad5.png)
inversion. Since this swaps 
and 
we have 
are swapped, so the circle with diameter 
is fixed, and 
gets sent to the foot from 
to 
is a symmedian.
with radius 
.
and 
themselves and same with 
and 
and the circle with diameter 
,
and obviously is the midpoint. 
be the midpoint of 
be the top point of 
,
collinear.
,
is cyclic.![\[\measuredangle MAD = \measuredangle MND = \measuredangle DAF. \quad \blacksquare\]](http://latex.artofproblemsolving.com/b/7/1/b7149517ba1ca4659451f0df62c429fecc7d338d.png)
which means that 
is fixed under this inversion. Now 
and thus, 
.
is simply the midpoint of 
be 
.
must lie on line 
,
;
;
is the midpoint of arc 
is nothing but the midpoint of 
is the reflection of the median about the angle bisector, i.e. the symmedian. 
be the antipode of 
,
.
,
,
.
denote the circle with diameter 
.
at 
and 
.
,
.
and 
at its midpoint 
with center 
passes through 
Take 
so 
and 
collinear. DDIT on 
means that 
and 
are isogonal from existing isogonal pairs 
and 
.
concur clearly. This is as 
is projected to 
.
is harmonic conj of 
are collinear and 
is cyclic. Note that 
so 
are also collinear and we get:![\[
\angle FAD=\angle FAE = \angle FNE = \angle DNM=\angle DAM
\]](http://latex.artofproblemsolving.com/0/a/b/0ab0704c1156f3fab3189527e5b838287d79b6d2.png)
and we're done since 
