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be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
, 
, 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
be positive real numbers so that 
. Prove that![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
, there exist positive real numbers 
, 
and 
such that 
, 
and 
. The inequality then reduces to![\[ \left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1. \]](http://latex.artofproblemsolving.com/d/a/6/da6f9bad78d4d92814b774a070ec5a199546e61a.png)
, 
and 
. The inequality then reduces to ![\[ 8pqr \le (p+q)(q+r)(r+p) \]](http://latex.artofproblemsolving.com/a/0/e/a0e912dc9dde61f92dd1e0d75fcdd321ecf6e623.png)
which is obvious from the AM-GM since at most one of 
can be negaitve..
so we have to prove this
then 
and 
an others then it ia sides of triangle so take
so from it follows. (we will look next step)
. Since 
are positive so
![\[8pqr \le (p+q)(q+r)(r+p)\]](http://latex.artofproblemsolving.com/e/3/9/e39bf0011b2a9ce9a212ac340670d747d32e60fb.png)
which is obvious from the AM-GM since at most one of can be negaitve..
are positive by definition. can be negative. If one of the numbers is negative, then it is trivial since the left side would be negative and the right side positive.
for example? If not, you can't appy AM-GM. There is a solution in that sense, but not as trivial as the one posted there.
,such that 
P
,
and multiply each factor
and so on. Multiply the product of denominators on both sides of the inequality and expand the 
and simplify a bit. The resulting inequality is just a case of Schur's.
We wish to show that 
or 
Expanding gives Schur's inequality, which is true.
hours
such that:![\[
a = \frac{x}{y}, \quad b = \frac{y}{z}, \quad c = \frac{z}{x}.
\]](http://latex.artofproblemsolving.com/4/6/4/464af3e2af4fb20cab9fae3d17ad0ce7dfadd50c.png)
Thus, the given inequality transforms into:![\[
(x + y - z)(-x + y + z)(x - y + z) \leq xyz.
\]](http://latex.artofproblemsolving.com/9/b/9/9b9cb9788281fec4f7c4b6dd1a5c81cf2701ff12.png)
Expanding this, we obtain:![\[
x^3 + y^3 + z^3 + 3xyz \geq x^2(y + z) + y^2(x + z) + z^2(x + y).
\]](http://latex.artofproblemsolving.com/b/7/3/b73dc791b148ac9993bad4f8ea7f4e5250c6ba15.png)
which is trivial by Schur's inequality
is almost essential to the problem. This basically simplifies out to 
.
,
,
and 
WLOG then we have 
which fails since x,y,z all must be positive. But if 
then the left hand side must be non-positive, and hence the negatives case is considered. We move on by assuming the Left Hand Side is positive.
,
,
.
are side lengths of a triangle. Write 
through ravi substitutions (yay). This becomes 
and therefore we take AM-GM on individual factors in the RHS which finishes the problem. 
for positive reals ![\[
(x - y + z)(y - z + x)(z - x + y) \le xyz.
\]](http://latex.artofproblemsolving.com/7/6/4/76461f23eb9524e2107325ece81d4ba2d8e26750.png)
Now, notice that if 
,
and 
for positive reals 
.![\[
8def \le (d+e)(e+f)(d+f).
\]](http://latex.artofproblemsolving.com/a/4/e/a4e0db3a479faef7aca7659bca20159cce238b7f.png)
Finally, by AM-GM, we have that 
,
and 
.
,
.![\[\frac{(x+y-z)(-x+y+z)(x-y+z)}{xyz} \le 1 \iff \sum_{\text{sym}} x^2y \le x^3+y^3+z^3+3xyz.\]](http://latex.artofproblemsolving.com/0/f/a/0fa9bb71f48dc8bdd60444915d3dcc99ef8f186a.png)
But this is Schurs with 
.
Hence, we have 
By expanding, we have that 
This follows from Schur's.
.
which is Schur's inequality for 
,
such that
Substituting into the original problem statement, we have to prove that
or equivalently, that
where we derive the second inequality by multiplying the first by 
.
,
,
.
(and all other cyclic variations) and we must prove that
which immediately follows from performing AM-GM three times over the pairs of any two of 
.![\[ (-x+y+z)(x-y+z)(x+y-z) \le xyz \]](http://latex.artofproblemsolving.com/4/2/2/4229cba5ae702b8162a925b098cfdc1c6c3639dc.png)
which when expanded is equivalent to Schur's inequality.