AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, 
, 
and 
positive integers such that:
+ 
+ 
= 
,
,
be real numbers different from 0 for which 
+ 
+ 
= 0 holds+)(+)(+)≠ 0
= + + and 
= 
+ 
+ 
. Prove that numbers and are both positive or both negative.
and such that 
+ 
+ 2 is a perfect square.
and be two positive integer numbers such that 
.Prove that 
divides 
be a positive integer. A board with a format 
is divided in equal squares.Determine all integers ≥3 such that the board can be covered in 
(or 
) pieces so that there is exactly one empty square in each row and each column.
.The Incircle touches sides 
and 
in 
and 
,respectivly. Lines CI and EF intersect at 
. The point 
≠ is on the line AI so that 
=
.If 
is the foot of the altitude from 
in triangle 
,prove that points , and are colinear.
be an acute scalene triangle and let 
be a point in its interior. Let 
, 
, 
be projections of onto triangle sides 
, 
, , respectively. Find the locus of points such that 
, 
, 
are concurrent and 
.
for which
is a prime number
such that there exists a sequence of positive real numbers 
, 
, with the property that for every it holds that:![\[
a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
\]](http://latex.artofproblemsolving.com/9/b/7/9b73deb46032844b324288b02470ff0623ac0ff1.png)
Proposed by Pavle Martinović
, let 
be the point which minimizes the sum of squares of distances from the sides of the triangle. Let 
the projections of on the sides of the triangle . Show that is the barycenter of 
.
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
, 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
, 
. Then the inequality reduces to![\[\frac{(x+y-z)(-x+y+z)(x-y+z)}{xyz} \le 1 \iff \sum_{\text{sym}} x^2y \le x^3+y^3+z^3+3xyz.\]](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.
, 
, 
then the inequality rewrites as ![\[ (-x+y+z)(x-y+z)(x+y-z) \le xyz \]](http://latex.artofproblemsolving.com/4/2/2/4229cba5ae702b8162a925b098cfdc1c6c3639dc.png)
which when expanded is equivalent to Schur's inequality.
and 
Prove that
.![\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]](http://latex.artofproblemsolving.com/b/3/2/b32e0bcba967ca847e878ac186c3168cd6c5fc66.png)
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)
such that ![\[ \left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1. \]](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 
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 
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.
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 
.
.
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 
.