[/quote]
,
,
be positive real numbers, such that 
.![\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]](http://latex.artofproblemsolving.com/b/e/5/be5819a67c3cd78f2dea35fdccf48688c720ce3c.png)
When does the equality hold?
be an acute triangle with orthocentre 
and circumcircle 
.
be the point
on 
,
and 
at 
,
intersect the circumcircle of triangle 
again at 
.
at 
.
and 
intersect on 
numbers on circle, and no one number is divided by other. In same time for all numbers we make next operation:
are two neighbors 
number 
and erase 
we can receive ?
numbers 
then after operation we receive circle with numbers 
.
.![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality.
?
,
,
,
.
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For 
represents 
.
are non-negative real numbers and 
Prove that
the equality hold for 
or 
or any
be positive real numbers. Prove 
Prove that
,
.
.
such that for all positive real numbers 
and 
:
and 
.![\[ \frac{a^2+1}{a+1}+\frac{b^2+1}{b+1}+\frac{c^2+1}{c+1} \geq 3\]](http://latex.artofproblemsolving.com/7/d/2/7d2e34116452912cd3da2925312ad500bb976f15.png)
be positive real numbers such that 
.![\[ \frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leq 0. \]](http://latex.artofproblemsolving.com/e/a/1/ea1f56cc594862673040411c3ad322cb00cfe904.png)
Y by Adventure10, Mango247, ehuseyinyigit
Some checkers placed on an 
checkerboard satisfy the following conditions:
(a) every square that does not contain a checker shares a side with one that does;
(b) given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares, such that every two consecutive squares of the sequence share a side.
Prove that at least
checkers have been placed on the board.
checkerboard satisfy the following conditions:(a) every square that does not contain a checker shares a side with one that does;
(b) given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares, such that every two consecutive squares of the sequence share a side.
Prove that at least
checkers have been placed on the board.
units available, where 
is the number of checkers. For each checker X with just two neighbors A, B: place the two ghost-checkers for X on the two squares adjacent to X different from A and B. For each checker with three neighbors, place one of its ghosts on the fourth adjacent square and keep the other. For each checker with four neighbors, keep both of its ghosts.
,
checkers, we have turned on at most 
checkers. Since there are 
squares on the board, we need 
or 
.
,
.
(but slightly larger due to edge effects).
checkers on the board (and 
empty squares), so from (b), the checkers form a connected graph 
.
.
,
.![\[n^2-f(n) \le Y\le4f(n)-2X,\]](http://latex.artofproblemsolving.com/9/c/9/9c909775e04d0a06c9508bf69e338d9faede7b30.png)
so![\[5f(n)\ge n^2+2X\ge n^2+2f(n)-2\implies f(n)\ge\frac{n^2-2}{3},\]](http://latex.artofproblemsolving.com/e/8/6/e8646f5549102e0cd19ade247788912995656147.png)
as desired.
squares, as each checker individually covers three squares, that is the two adjacent squares not covered by another checker as well as the square that the checker itself is on, while the first and last squares additionally cover another square each. Note that no square is covered twice. It follows that 
,
.
by 
,
checkers, we have that it will take greater than 
.
and the denominator of three in the fraction. The three suggested looking at how many new squares each new checker in a "snake" of checkers would cover, while the 
of checkers where the 
edges of the tree are given by orthogonal adjacency. By condition (a) we have ![\[ \sum_{v \in V} (4-\deg v) \ge n^2 - |V| \]](http://latex.artofproblemsolving.com/7/5/7/757abf146f7425759345f79a24ca6876bac7a466.png)
and since 
we get ![\[ 4|V| - \left( 2|V|-2 \right) \ge n^2 - |V| \]](http://latex.artofproblemsolving.com/a/7/7/a77a86e75e71be146e7f0a600e670f434bcb9245.png)
which implies 
.
and we are done
squares, and each successive one must add only 
be the number of checkers. We say that a cell is good if it either contains a checker or is adjacent to a cell that contains a checker. Since the order we place them in does not matter, we placed them in an order such that each checker after the first is placed next to an existing checker (this is possible due to the connected component condition).
,
(the number of empty squares). On the other hand, the number of pairs of adjacent squares, both filled, is at least 
.
pairs of adjacent squares, at least one of which is filled. Therefore, we have the inequality 
as required.
as each new block after the original one adds at most 
to the perimeter of the filled squares. The minimum perimeter is 
,
![\[1+\frac{n^2-5}{3},\]](http://latex.artofproblemsolving.com/3/2/9/32993f5563d49695dcb6fc04e90ef0a94f339db2.png)
or ![\[ 3k+2 \leq n^2 \implies k \geq \frac{n^2-2}{3} \]](http://latex.artofproblemsolving.com/5/a/a/5aa15265602953636076d6efac2ea40b7d77e8e8.png)
completing the proof. 
squares because of the first condition. Thus,![\[3k+2 \ge n^2 \iff k \ge \frac{n^2-2}{3}. \ \square\]](http://latex.artofproblemsolving.com/c/f/6/cf6c0428fed9892d6cde124c04ed14eea6e629f1.png)
be the number of white cells and let 
be the number of black cells. Then each white cell has at least one black neighbor.
edges. Therefore, the black cells can satisfy at most 
white cells, so 
.
,
be the set of squares with and without a checker, respectively. A upper bound for 
is ![\[\sum_{a \in A} (4-\deg a) = 4|A| - \sum_{a \in A} (\deg a)\]](http://latex.artofproblemsolving.com/4/4/e/44ee2c2b32119a0dcbb221a1fddeddf0b81829d4.png)
.
edges, and thus![\[4|A| - \sum_{a \in A} (\deg a) \ge 4|A| - 2(|A|-1) \ge |B| = n^2 - |A| \implies |A| \ge \frac{n^2-2}{3}. \quad \blacksquare\]](http://latex.artofproblemsolving.com/b/d/b/bdbada60cbe2c0bbcb8e80b2407e32607acd4ad1.png)
be the number of checkers next to square 
,
goes up by 
,![\[\sum_{i=1}^{n^2}C_i\geq 2a-2+n^2-a = n^2-2+a.\]](http://latex.artofproblemsolving.com/e/9/2/e92cdc34c0285c1459263edf431385afd95a2a6d.png)
However, notice that each checker is counted at most four times, because it has at most 
surrounding squares. This gives us ![\[4a\geq\sum_{i=1}^{n^2}C_i\geq n^2-2+a,\]](http://latex.artofproblemsolving.com/9/7/3/973228b48b70e04a83dc386505fb20611ce40a87.png)
giving 
,
,
checkers have at most 
.
.
.
,
.
be the number of black neighbors of 
over all squares so that we can get a lower bound on 
.
.
be a graph with its vertices as the ![\[\sum_{s\text{ black}} f(s) = \sum_{v \in V} \text{deg}(v) = 2|E| \geq 2k - 2,\]](http://latex.artofproblemsolving.com/0/c/c/0cc9a9977a334ab7f84b9e641a056a16fbb88d65.png)
where the last inequality follows since 
.
is counted 
.![\[4k \geq n^2 + k - 2 \implies k \geq \frac{n^2 - 2}{3}\]](http://latex.artofproblemsolving.com/b/7/5/b7574b0b72819d89a18580af6cb92b9e6b511ac3.png)
as desired. 
,
is never divisible by 
denote the set of edges in this graph.
because each square with a checker on it corresponding to the vertex 
in the graph has 
neighboring squares without checkers on them. Also, a lower-bound for this quality is the number of squares without checkers which is 
because each square without a checker on it must be adjacent to at least one square with a checker on it (condition (a)). Thus, it follows that 
so we can apply the Handshaking Lemma to get
Condition (b) is equivalent to the graph being connected, so we know that 
Hence, it follows that
Therefore, we get
as desired. 
squares we need one of their sides to touch a square with a checker. Therefore, the perimeter of our 
(this is from the second condition and that some squares touch sides of the grid). Therefore 
done
If 
we have at least 
empty squares adjacent to a filled square. Thus 
QED
or 
rectangle such that it contains exactly one checker. Count 
.
.
,
.
we have that 
By 
we have that 
Finally, since one checker can border only 4 squares, we have that 
Thus, we have that 
