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.
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut . such that, on the -th move, Jumpy swaps some walnuts 
and 
such that 
.
kawaii if it satisfies 
(
is the number of positive factors of , while 
is the number of integers not more than that are relatively prime with ). Find all that is kawaii.
. Find the minimum and maximum values of 
+ 
+ 
such that the equation![\[f(xf(x)+y) = f(y) + x^2\]](http://latex.artofproblemsolving.com/2/0/a/20a245460634f0455fc703d7ce41c475c1ae4970.png)
holds for all rational numbers 
and 
.
denotes the set of rational numbers.
a point in the space and 
the centroid of a tetrahedron 
. Prove that:
and be positive integers. If 
is divisible by 
for every positive integer , prove that 
.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
, 
, 
be nonzero real numbers such that 
and 
. Find the value of
.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be a function satisfying
for all integers and . Show that there exist positive integers 
and such that 
for all integers .
for 
for positive reals .
for positve reals .
![\[ a^4 + b^4 = 4ab - 2\]](http://latex.artofproblemsolving.com/1/5/6/156f7be5ac3d886491d230194b9e2c6400797ef3.png)
in positive reals.
for for positive reals .
for for positive reals .
when 
for for positive reals . for positve reals .
for 
positive integers.
are positive reals with 
. Prove![\[ \sum_{cyc} \frac{1}{a^3(b+c)} \ge \frac{3}2\]](http://latex.artofproblemsolving.com/3/2/7/3274c741499c72d0164802a7864d8e5577fc81b4.png)
are non negative reals such that 
. Prove![\[ a^3+b^3+c^3+6abc \ge \frac{1} 4 \]](http://latex.artofproblemsolving.com/a/9/a/a9a48560ffaa2cb72ed3688af8402e0b56e52a69.png)
for for positive reals . for positve reals .
![\[\sum^{10}_{x=2} \dfrac{2}{x(x^2-1)}\]](http://latex.artofproblemsolving.com/a/2/7/a270f612ff7d9d39b3c90e668165f6bf9c1358d1.png)
.
and 
.
.
,
.
.
,
and 
.
.
cancels out all terms, 
.
.
which is true from applying 6-variable AM-GM.
,
because rearranging gives 
which is true from trivial inequality. So we want them all to be equal, which is only possible when 
(because of AM-GM"s equality condition. Then we must have 
.
and 
.![$[(x+y)(y+z)(z+x)]^2\ge [8xyz]^2\Rightarrow (x+y)(y+z)(z+x)\ge 8xyz$](http://latex.artofproblemsolving.com/4/3/6/43647338b5e7ba424736e600f0f592181560bfe0.png)
by taking positive square roots.
for positive a,b,c
or 
when ![\[\left(\dfrac{a^3+b^3+1}{ab}\right)\left(\dfrac{b^3+c^3+1}{bc}\right)\left(\dfrac{c^3+a^3+1}{ca}\right).\]](http://latex.artofproblemsolving.com/7/4/6/7463c2e4cfc9d3e31740a16a27e958e5c3d55cec.png)
By AM-GM, we have ![$\dfrac{a^3+b^3+1}{ab}\geq\dfrac{3\sqrt[3]{a^3b^3}}{ab}=3$](http://latex.artofproblemsolving.com/1/7/5/175dd3c56948cef8fd9fa1d05b8c20262f2ac9ee.png)
.
.
,![\[ x + y \ge 2\sqrt{xy} \\ y+z \ge 2\sqrt{yz} \\ z+x \ge 2\sqrt{zx} \]](http://latex.artofproblemsolving.com/e/3/8/e380b0a520f67702472eb8da66a079672870accb.png)
Multiplying the three inequalities gives 
from four-variable AM-GM. Therefore 
.
or 
,
.
where 
by five variable AM-GM.![\[\dfrac{x+x+x+x+\dfrac{1}{x^4}}{5} \ge \sqrt[5]{x \cdot x \cdot x \cdot x \cdot \dfrac{1}{x^4}} = 1\]](http://latex.artofproblemsolving.com/c/4/b/c4bbdc916ad6dddf472840539bb11d262d89ea10.png)
so 
.
.
gives ![\[\sum_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{x^2}{y+z}.\]](http://latex.artofproblemsolving.com/5/d/8/5d826cf943840687ca0b641ce461693f0a8a1bdb.png)
From AM-GM we have 
and ![$x+y+z\geq 3\sqrt[3]{abc}=3$](http://latex.artofproblemsolving.com/3/7/5/375d6e0603a4db834ee98a734695342c534eb696.png)
,![\[\sum_{cyc}\frac{x^2}{y+z}\geq \sum_{cyc}\left(x-\frac{y+z}{4}\right)=\frac{x+y+z}{2}\geq \frac{3}{2},\]](http://latex.artofproblemsolving.com/f/3/4/f34ff0447752d3366a262073882e04fe1bd5e38c.png)
as desired. From the nature of AM-GM, equality holds at 
.
which is weaker than and immediately follows from Schur. For equality, from Schur we need 
,
are 
and the other is 
.