90. Nice and difficult inequality in ABC (own).
by Virgil Nicula, Aug 26, 2010, 3:45 PM
PP1. 
for which 
with equality iff 
and 
.
Remark. Inequality
is equivalently with 
.
Can use
. Particular case. 
, i.e. 
.
Proof. Denote second intersection
of 
-symmedian with circumcircle 
and 
. Observe that
. Apply remarkable inequality 
to triangle 
:
![$\frac {2bc}{a}\cdot \left[\frac {k}{k^2+1}+\frac {1}{k(k^2+1)}\right]+\frac {k+1}{k}\cdot b\le 3R\sqrt 3$](//latex.artofproblemsolving.com/1/1/4/114d16636615496f2fcf489782548e4897d085fa.png)
.
PP2 (an equivalent enunciation). Let
be a triangle with the circumcircle . The -symmedian cut 
in and meet again 
in
. Denote the midpoint 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, 
, 
and 
, 
. Prove that the following relations:
and 
.
(the quadrilateral 
is harmonically) and 
.
. Particular case. 
.
Remark. I used the notation
- the tangent line to the circle at the point 
.
for which 
with equality iff 
and 
.Remark. Inequality
is equivalently with 
.Can use
. Particular case. 
, i.e. 
.Proof. Denote second intersection
of 
-symmedian with circumcircle 
and 
. Observe that
. Apply remarkable inequality 
to triangle 
:
![$\frac {2bc}{a}\cdot \left[\frac {k}{k^2+1}+\frac {1}{k(k^2+1)}\right]+\frac {k+1}{k}\cdot b\le 3R\sqrt 3$](http://latex.artofproblemsolving.com/1/1/4/114d16636615496f2fcf489782548e4897d085fa.png)
.PP2 (an equivalent enunciation). Let
be a triangle with the circumcircle . The -symmedian cut 
in and meet again 
in
. Denote the midpoint 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, 
, 
and 
, 
. Prove that the following relations:
and 
.
(the quadrilateral 
is harmonically) and 
.
. Particular case. 
.Remark. I used the notation
- the tangent line to the circle at the point 
.This post has been edited 41 times. Last edited by Virgil Nicula, Nov 23, 2015, 2:03 PM
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