AoPS Academy
. The length of the first jump is 
, the second 
, the third 
, and so on. The length of 
jump is equal to 
. The flea decides whether to jump left or right on its own. Is it possible that sooner or later the flee will have been on every natural point, perhaps having visited some of the points more than once?
over all pairs of positive integers 
.
lamps 
arranged in a circle in that order. At any given time, each lamp is either on or off. Every second, each lamp undergoes a change according to the following rule:
, if 
have the same state in the previous second, then is off right now. (Indices taken mod .) is on right now.
which is on. Prove that for infinitely many integers all the lamps will be off eventually, after a finite amount of time.
and 
, let us consider the equation
[list=a]
and 
, find the least positive integer 
satisfying this equation. and , there exist infinitely many positive integers satisfying this equation.
denotes the greatest common divisor of positive integers 
and .)
for which there exist real numbers 
satisfying 
, 
and
for 
.
such that 
holds for all positive rational numbers 
.
and 
. Prove that
Equality holds when 
Equality holds when 
and are prime-related if 
or 
for some prime 
. Find all positive integers , such that has at least three divisors, and all the divisors can be arranged without repetition in a circle so that any two adjacent divisors are prime-related.
and are included as divisors.
is chosen inside a convex quadrilateral 
. Could it happen that
and cyclic variants, then we can rewrite as 
. Observe at most one of the cyclic terms can be negative, since if the variable being negated is not maximized we can easily see the term is positive. If one term is negative, we instantly win by sign, otherwise we see that 
and cyclic variants so 
are sides of a triangle, set 
with 
and cyclic variants, then we are reduced to proving 
, which is obvious by AM-GM on each term on the RHS.
with 
. Then we have![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1 \iff (x+y-z)(x-y+z)(-x+y+z) \le xyz.
\]](http://latex.artofproblemsolving.com/d/b/9/db9b2e605e5c3b3250ee9ff79f947e076c2af769.png)
Now we have two cases:
which are negative is at least 
.
then adding yields a contradiction..
Case 3: All of the terms are nonnegative.
. with 
. Then the inequality reduces to 
which is true by AM-GM.
with 
.
, the above expression is equivalent to:
Now by using the substitution 
we can homogenize the above inequality. Thus we have to show that,
which is trivial by Schur's and hence, we are done. 
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.
, there exist positive real numbers 
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
be positive real numbers so 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)
and 
such that 
,
and 
.![\[ \left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1. \]](http://latex.artofproblemsolving.com/d/a/6/da6f9bad78d4d92814b774a070ec5a199546e61a.png)
,
and 
.![\[ 8pqr \le (p+q)(q+r)(r+p) \]](http://latex.artofproblemsolving.com/a/0/e/a0e912dc9dde61f92dd1e0d75fcdd321ecf6e623.png)
which is obvious from the AM-GM since at most one of 
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 
are positive by definition.
and multiply each factor
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.
,
,
.![\[\prod_{\text{cyc}} \frac{x+z-y}{y} \le 1\]](http://latex.artofproblemsolving.com/0/9/f/09f3e716c14f66f17353de5ce3d6384ee4b1e749.png)
Expanding, it suffices to prove that![\[\sum_{\text{cyc}} x^3 + 3xyz \ge \sum_{\text{sym}} x^2y\]](http://latex.artofproblemsolving.com/5/b/8/5b8c481426bfdb576e06cb4341e589c78715b91d.png)
which is true by Schur.
)
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.
,
,
.
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 
.