AoPS Academy
, the points 
and 
are taken such that the point lies between the points 
and , and 
. On segments 
and 
, points 
and 
are taken so that 
and 
. Prove that 
.
and points and on sides 
and 
, respectively, such that 
. Let 
and 
be intersection points of lines 
and 
with circumcircle 
of triangle 
, and 
and 
intersection points of lines 
and 
with side 
. If 
, prove that circle contains intersection point of angle bisector of 
with
be an integer. Let 
be 
positive real numbers satisfying the following conditions:
we have 
and,
the numbers 
form a permutation of 
, determine the maximum value of the sum 
This can obviously be achieved by letting all entries of the matrix be equal to Using the condition, we could rewrite the inequality as 
. When 
this is just 
, which is obviously true - we have equality, in fact., and we will prove it for 
Denote 
for any 
, so that 
for all 
![\begin{align*}
\sum_{j=2}^{n+1}\frac{1}{S'_j}-\sum_{j=1}^{n+1}\frac{1}{S_j} &= \left[\sum_{j\ge 2}\left(\frac{1}{S'_j}-\frac{1}{S_j}\right)\right]-\frac{1}{S_1}\\
&= \left[\sum_{j\ge 2}\frac{\left(\frac{1}{S'_j}\right)^2}{a_{1j}+\frac{1}{S'_j}}\right]-\frac{1}{S_1} \\
&\ge\frac{\left(\sum_{j\ge 2}\frac{1}{S'_j}\right)^2}{S_1-1+\left(\sum_{j\ge 2}\frac{1}{S'_j}\right)}-\frac{1}{S_1},
\end{align*}](http://latex.artofproblemsolving.com/1/9/f/19fee369e779af33605df6630eb159877ca4ffbc.png)
and 
Note that 
by the induction hypothesis (too see why, just eliminate the first row of the upper triangular matrix). Thus,
, thus establishing the induction step. This ends our proof.
