AoPS Academy
, 
, 
, ... be an infinite sequence of real numbers satisfying the equation 
for all 
, where and are two different positive reals., , , ... be bounded?
be two positive sequences defined by 
and![\[ \begin{cases} {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\ x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \]](http://latex.artofproblemsolving.com/2/0/1/2012dc6ff3a41478547cea4aa9b6bccbe17a3623.png)
for all 
.
be an infinite sequence of positive integers. Prove that there exists a unique integer 
such that![\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]](http://latex.artofproblemsolving.com/9/4/f/94fc5a51b7e68588123c5b527fe75183bc4c4937.png)
, define a sequence of triples 
by
.Show that each of the sequences 
converges to a limit and find these limits.
, define a sequence of triples by
,.Show that each of these sequences converges to a limit and find these limits.
are the three positive sidelengths or a triangle, we can rather easily show that :
are the sides of a new triangle with same area
, 
and 
are decreasing sequences
have same limits, sides of an equilateral triangle of same area
![$l=\sqrt[4]{\frac{(x_0+y_0+z_0)(x_0+y_0-z_0)(x_0-y_0+z_0)(-x_0+y_0+z_0)}3}$](http://latex.artofproblemsolving.com/7/4/9/749d0a7955dd294a78fcf172aa31ed1225df1dc7.png)
, this formula gives 
. Also, the thing under the radical suspiciously looks like the product of length of tangent from a vertex and some other thing. This motivates the introduction of a triangle with sides 
. Usually we need some kind of invariant in multivariable limits. So, we try the squared area of the triangle since it is a polynomial in , more precisely, 
. This surprisingly comes out to be invariant. Also to show the convergence, it seems better to choose the difference between squares.
