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 rightways, Mr.C, Adventure10, Mango247, Rounak_iitr
In 
, let 
and 
be on segments 
and 
respectively, and let 
and 
intersect at 
. Define 
on segments 
such that 
and its cyclic equivalents. Let the incircle of touch 
at 
respectively; let the incircle of 
touch 
at 
respectively.
Prove that concyclic implies 
concyclic.
, let 
and 
be on segments 
and 
respectively, and let 
and 
intersect at 
. Define 
on segments 
such that 
and its cyclic equivalents. Let the incircle of touch 
at 
respectively; let the incircle of 
touch 
at 
respectively.Prove that
concyclic implies 
concyclic.
.
.
.
,
,
.
and so on, 
,
and so on, which implies 
,
is tangential.
be the tangency points of the inscribed circle of 
respectively.
,
.
,
.
,
concyclic, and we are done.
,
,
and 
,
, 
,
,
,
centred at 
,
across line 
is an involution and 
,
,
,![[asy]
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[/asy]](http://latex.artofproblemsolving.com/a/f/7/af77cb8ca5ce621282933f5718fbd48f15bda136.png)
be the intersection of the angle bisector of 
and 
,
.
Hence 
is the angle bisector of 
,
,
are concyclic. Similarly 
are concyclic.
where 
is the angle bisector of 
and 
Since they are equal,
Rearraning gives
Let 
be intersection of the angle bisector of 
and 
.
similarly. Then From 
we have 
for some 
.
,
and 
,
,
is impossible. This implies 
so 
,
,
and 
so we are done. ![[asy]
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[/asy]](http://latex.artofproblemsolving.com/e/5/1/e5184b4bad63559cdd9aafb51cb2cd6841da0d99.png)
Indeed, 
,
Notice that 
is the touching point of the 
-
,
equals 
while the right hand side of 
.
be the intersection of 
and 
.
.
are collinear. As a result
hence 
