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.
, 
, 
be positive real numbers such that 
. Prove that ![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
and be positive real numbers satisfying 
and 
Prove that![\[\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.\]](http://latex.artofproblemsolving.com/0/a/b/0ab4b4e6c28a1ae67788525642661dbcd04d0bd8.png)
such that ![\[ f(x^2 + yf(z)) = xf(x) + zf(y) , \]](http://latex.artofproblemsolving.com/2/b/a/2baee1d8464992fc38a16561b48b18f4b35807bd.png)
for all 
.
and in 
steps and has to reach again. But in between one of the s steps and can't go . Find probability. For example 
is valid but 
is not valid.
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
such for any :
.
satisfy,
Where {
} is a permutation of . Prove that 
be an integer. We are given a balance and 
weights of weight 
. We are to place each of the weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.
to satisfy that the equation 
has at least 1 roots 
of positive integers is called central if for every positive integer , the arithmetic mean of the first 
terms of the sequence is equal to .
, 
, 
, 
of positive integers such that for every central sequence 
, 
, 
, , there are infinitely many positive integers with 
.
of the equation![\[n^3=p^2-p-1\]](http://latex.artofproblemsolving.com/f/f/8/ff88093bb0aa93afcfdff3ae25149ec5fbacb61f.png)
is a prime number and is an integer
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of , the maximum number of good cells over all possible starting configurations.
denote the lengths of the sides 
and the diagonals 
of a cyclic quadrilateral 
respectively. Prove that 
. Let 
be the images of 
. Writing Stewart's theorem for 
and the cevian 
we find![\[\frac{y}{x} = \frac{bc+da}{cd+ab}.\]](http://latex.artofproblemsolving.com/7/2/d/72d657faa7b3904c5844db385ca1cdc52e576414.png)
Combining this with Ptolemy's theorem 
we find![\[x^2 = \frac{(ac+bd)(bc+da)}{(ab+cd)}, \qquad y^2 = \frac{(ac+bd)(ab+cd)}{(bc+da)}.\]](http://latex.artofproblemsolving.com/8/6/8/8689d09e1e10dfe7e89d7e561877f279bff82d3b.png)
Thus our inequality becomes![\[\left(\frac{a+c}{ac} \right)^2 + \left(\frac{b+d}{bd} \right)^2 \ge \frac{8}{ac+bd}\left(\frac{bc+da}{ab+cd} + \frac{ab+cd}{bc+da}\right)\]](http://latex.artofproblemsolving.com/d/6/a/d6aac7cdf4805ed1ab05b2e6304237b96aa236a0.png)
which is true by AM-GM:![\[LHS \ge \frac{2(a+c)(b+d)}{abcd} \ge \frac{4}{ac+bd}\cdot \frac{ab+cd+bc+da}{\sqrt{abcd}}\ge RHS.\]](http://latex.artofproblemsolving.com/9/4/8/9482c61b06b099f835f2536822c8fba6ce2a935f.png)
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