be a positive integer. Given a sequence of nonnegative real numbers 
we define the transformed sequence 
as follows: the number 
is the greatest possible value of the average of consecutive terms of the sequence that contain 
.
is 
.
,
is less than or equal to 
.
holds.
be positive real numbers. Prove that:![$$\sqrt[3]{a^3+b^3}+\sqrt[3]{b^3+c^3}+\sqrt[3]{c^3+a^3}\geq \sqrt[3]{2}(a+b+c)$$](http://latex.artofproblemsolving.com/2/6/e/26ebcdae35153bee73abbcfaf4a767d3a32b8912.png)
be all the positive integers smaller than 
such that
for all 
.
is written on the board, with 
linear factors on each side. What is the least possible value of 
for which it is possible to erase exactly 
linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
Y by Adventure10, Mango247
Let 
be a parallelogram. Let 
be a point on the side 
and 
be a point on the side 
such that the segments 
and 
have equal lengths and are non-zero. The lines 
and 
meet at 
.
Prove that the line
is the bisector of the angle 
.
Alternative formulation. Let be a parallelogram. Let and be points on the sides and , respectively, such that 
. The lines and intersect at a point .
Prove that the point lies on the bisector of the angle .
be a parallelogram. Let 
be a point on the side 
and 
be a point on the side 
such that the segments 
and 
have equal lengths and are non-zero. The lines 
and 
meet at 
.Prove that the line
is the bisector of the angle 
.Alternative formulation. Let
be a parallelogram. Let and be points on the sides and , respectively, such that 
. The lines and intersect at a point .Prove that the point
lies on the bisector of the angle .
be the intersection point between 
in 
,
is the interior angle bisector if and only if (the bisector theorem)![\[ \frac{ AP}{PQ} = \frac{AD}{DQ} \quad (1) \]](http://latex.artofproblemsolving.com/c/f/c/cfcd05beebd0f54b1ada10edf7c00796f3c94625.png)
and 
are similar, thus![\[ \frac{AP}{PQ} = \frac{AM}{CQ} = \frac{CN} {CQ } = \frac{AD}{DQ} \]](http://latex.artofproblemsolving.com/1/c/2/1c2a067550632cc416e2590dab30e1e5c6e303dd.png)
where the last equality is derived from Thales for the triangle 
and the parallel 
, and when I had the limit I represented Romania 
is homographic between the lines 
(with points 
considered as origins on these lines respectively). The locus of the point 
and the line at infinity. These cases are solved easily and we find that 
and 
.
.
.
,
.![\[ \frac{XQ}{QC}=\frac{XA}{CN}=\frac{XA}{AM}=\frac{XD}{DC} \]](http://latex.artofproblemsolving.com/b/8/4/b843ad02fb6425fb152f70511b4946750fd99e63.png)
Now done by angle bisector theorem.
we show 
or since 
and 
we only need to prove 
.
(Points 
and 
Note that 
so 
