prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
satisfying 
is inscribed in circle 
.
denote its incenter and 
its 
denote the midpoint of arc 
.
meets 
.
at 
meets 
a second time at 
meets 
a second time at 
.
passes through the antipodal to 
that satisfy 
with 
,
and area 
.
with diameter 
,
and extend to meet 
and 
is the diameter of 
of 
,
of 
and 
Prove that
Let 
Prove that
Let 
Prove that
Let 
Prove that
.
lie on 
respectively, such that 
.
to meet the circumcircle of 
respectively. Let the circumcircles of 
meet at points 
.
to 
,
,![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
,
to meet the circumcircle of 
at 
,
.
Y by Adventure10, Mango247
Let 
be an acute triangle and let 
be the foot of the perpendicular from 
to side 
. Let 
be a point on segment 
. Lines 
and 
intersect sides 
and 
at 
and 
, respectively. Let 
and 
be the feet of the peroendiculars from and to , respectively. Show that
.
be an acute triangle and let 
be the foot of the perpendicular from 
to side 
. Let 
be a point on segment 
. Lines 
and 
intersect sides 
and 
at 
and 
, respectively. Let 
and 
be the feet of the peroendiculars from and to , respectively. Show that
.
- 
.
.
- 
.
.
is equivalent to
.
.
.
.
,
.
.
Since the cevians 
are concurrent in 
But since 
,
bisects 
Hence, 
,
and 
this can be easily done applying the generalized angle bisector theorem in an isosceles triangle.
and 
which implies also using the fact that 
that 
and the desired result follows 
that send 
is a parallelogram, also 
,
is a parallelogram
)
and 
)![\[\Rightarrow \frac{XJ}{XF}=\frac{JD}{KD}=\frac{EJ}{FK}\]](http://latex.artofproblemsolving.com/8/d/2/8d225d6ac40eb425666111655a667c87fd230996.png)
and 
concurrent at ![\[\frac{FK}{KD}=\frac{EJ}{JD}\]](http://latex.artofproblemsolving.com/f/7/e/f7e21e465d965341157e2e9322e9a9a359648586.png)
