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.
. Determine all functions 
such that![\[ xf(y) + yf(x) = (x+y) f(x^2+y^2) \]](http://latex.artofproblemsolving.com/0/c/3/0c3b10a82eb78d0dac62560b6736e755ecefe350.png)
and 
in 
.
a function satifying 
for all 
.
, then what is 
, the answer is in terms of 
.
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
and the recursive relation
for all natural number 
. Find the general form of 
.
of the set 
, define its colorfulness as the greatest natural number 
such that:
, 
, if 
, then 
.
? Determine how many such permutations have maximal colorfulness.
, 
, with 
, 
. Prove that there exists 
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 and , the notation 
represents 
. )
. The segment 
measures 40 units. Let 
be the orthocenter of triangle and 
its circumcenter. Let 
be the foot of the altitude from 
and 
the foot of the altitude from 
. Additionally, point divides the segment such that 
. If the perpendicular bisector of segment 
passes through point , calculate the area of quadrilateral 
.
be real number such that 
Prove that
Let 
be real number . Prove that
be a triangle. The points 
and 
lie on the segments 
and 
respectively, such that the lines 
and 
intersect in a common point. Prove that it is possible to choose two of the triangles 
and 
whose inradii sum up to at least the inradius of the triangle .
.But then when does equality hold's.
and 
positive real numbers. Prove that
.But then when does equality hold's.
we get equality in the given inequality.
and positive real numbers.Prove that
for all 
; hence 
with the last inequality following from AM-GM. Equality occurs when 
.
![$ P\in R[X]$](http://latex.artofproblemsolving.com/d/6/7/d67db7e6fb626de909cab78c68abbc61b5bcb303.png)
with all positive coefficients.Prove that![\[\frac{1+P(x_{1})^2}{1+P(x_{1}t(x_{1}))}+\frac{1+P(x_{2})^2}{1+P(x_{2}t(x_{2}))}+\cdots+\frac{1+P(x_{n})^2}{1+P(x_{n}t(x_{n}))}\geq n.\]](http://latex.artofproblemsolving.com/2/a/f/2af55eb6915b326649d608cf876d5d90f3dd63fc.png)
,where 
is a permutation of nubers 
.
So the equation turns into
where the last inequality follows from AM-GM and we are done.
-variable inequalities for some reason)
, i.e.
where 
.
such that 
for all 
, we have the inequality
Summing this over all choices of over all choices of 
yields the expansion of the desired inequality, so we are done. 
![$$\sum_{i=1}^n \frac{1+x_i^2}{1+x_ix_{i+1}} \geq n \sqrt[n]{\frac{\prod_{i=1}^n (1+x_i)^2}{\prod_{i=1}^n (1+x_ix_{i+1})}}.$$](http://latex.artofproblemsolving.com/f/e/8/fe87b4ea35058a9a923e298abc2a6491d682f6ae.png)
But just multiply 
cyclically to get the result.
![\[\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}} \ge n \sqrt[n]{\prod_{\text{cyc}} \frac{1+x_{1}^2}{1+x_{1}x_{2}}}\]](http://latex.artofproblemsolving.com/4/d/1/4d1e7ea77a75ded81b602a5e55013338431ce1a9.png)
![\[\prod_{\text{cyc}} \frac{1+x_{1}^2}{1+x_{1}x_{2}} \ge 1\]](http://latex.artofproblemsolving.com/9/b/9/9b98d5cc197d08e4d7f7c74d99b33b3c5bc77fe4.png)
![\[(1+x_i^2)(1+x_{i+1}^2) \ge (1+x_ix_{i+1})^2\]](http://latex.artofproblemsolving.com/4/8/9/489be11d4d3cbc6b5164afa1c391b6cc62809101.png)
which follows from multiplying the Cauchy Schwarz 
cyclically.
By AM-GM, we have that ![$$\text{LHS} \geq n \sqrt[n]{\prod_{i=1}^{n} \frac{1+x_1^2}{1+x_1x_2}},$$](http://latex.artofproblemsolving.com/4/7/5/475e925190e60bbeaa4d74e106d798074f22dc4a.png)
so it suffices to show that 
However, note that by Cauchy-Schwartz 
so multiplying all of these inequalities for 
and then taking the square root yields the desired result. QED
Squaring both sides, and matching, CS finishes.
.![\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]](http://latex.artofproblemsolving.com/b/3/2/b32e0bcba967ca847e878ac186c3168cd6c5fc66.png)
be real numbers such that 
.
and 
Prove that
Prove that
Let 
Prove that
positive real numbers. Prove that![\[\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n.\]](http://latex.artofproblemsolving.com/e/9/e/e9e73a14f4dfa6315e7562998a9394d371847139.png)
.![$$\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n\sqrt[n]{\frac{1+x_{1}^2x_2^2}{1+x_{1}x_{2}}\cdot\frac{1+x_{2}^2}{1+x_{2}x_{3}}\cdots\frac{1+x_{n}^2}{1+x_{n}x_{1}}}$$](http://latex.artofproblemsolving.com/c/d/c/cdc2bcd1410605a701c3d8ca3a64436709cd453d.png)
Now, by C-S, we have for all 
(indices are taken modulo 
.
whence we have our desired.
![$$\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+...+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n\sqrt[n]{\frac{1+x_{1}^2}{1+x_{1}x_{2}}\cdot \frac{1+x_{2}^2}{1+x_{2}x_{3}}\cdots\frac{1+x_{n}^2}{1+x_{n}x_{1}}}\geq n$$](http://latex.artofproblemsolving.com/3/7/3/37383bea956e7f43ea802cc267718e0079d3202b.png)
and 
I think this task inspired Russia MO
![\[\sum\frac{1}{1+x_ix_{i+1}}+\sum\frac{x_i^2}{1+x_ix_{i+1}}\ge n\]](http://latex.artofproblemsolving.com/b/b/6/bb66cd3658b269490fecddef29413eac6e802b08.png)
now apply titu we get:![\[LH \ge \frac{n^2}{n+\sum x_ix_{i+1}}+\frac{\sum x_i^2}{n+\sum x_ix_{i+1}}\ge^? n\]](http://latex.artofproblemsolving.com/b/d/b/bdbf1aaf1b8443068e45662252df1177e4918600.png)
which is obviously true because:![\[\sum \frac{x_i}{2}+\frac{x_{i+1}}{2} \ge \sum x_ix_{i+1}\]](http://latex.artofproblemsolving.com/7/1/a/71ae417ffebfb77b4d741fdba7ca35b8b0a78a59.png)
and 
show that ![\[\sum_{k=1}^n\frac{1+x_k^2}{1+x_kx_{k+1}}\ge n.\]](http://latex.artofproblemsolving.com/9/a/4/9a4243d17e497dbbb0c14f9695ff78e6cb7a1fcd.png)
![\[\sum_{k=1}^n\frac{1+x_k^2}{1+x_kx_{k+1}}\ge\frac12\left(\sum_{k=1}^n\frac{1+x_k^2}{1+x_kx_{k+1}}+\sum_{k=1}^n\frac{1+x_kx_{k+1}}{1+x_{k+1}^2}\right)\ge n.\]](http://latex.artofproblemsolving.com/8/a/a/8aa682a2fd15332173fdfef0a3fb92f3dae1e2e2.png)
such that for any two positive real numbers 
we have: ![\[\frac{1+x^2}{1+xy} \geq \frac{f(x)}{f(y)}\]](http://latex.artofproblemsolving.com/d/3/6/d36acf65cb08e14d8288ea04a1c348c8b7cf8775.png)
Observe that if we suceed in finding such a function 
,![\[[f(y)-f(x)]+x \cdot [xf(y)-yf(x)] \geq 0\]](http://latex.artofproblemsolving.com/e/0/8/e0854ee73ccf6e214e192b2cf2121b976142d4e9.png)
for all positive 
with 
,
and take limits for 
.![\[-f'(y) + yf(y) - y^2f'(y) \geq 0\]](http://latex.artofproblemsolving.com/8/d/7/8d79bc0c0a28f9c7e398415774969cddc10207e9.png)
or that ![\[\frac{y}{1+y^2}\geq \frac{f'(y)}{f(y)}\]](http://latex.artofproblemsolving.com/e/e/5/ee5a0b96b9ad24982d495ac4981d5aa2b494d98e.png)
Assuming equality in the above bound and solving for 
.
for all positive real numbers 
works and we're done.
![\begin{align*}
\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}} \geq n\sqrt[n]{\dfrac{\left(1+x_1^2\right)\left(1+x_2^2\right)\cdots \left(1+x_n^2\right)}{\left(1+x_1x_2\right)\left(1+x_2x_3\right)\cdots\left(1+x_nx_1\right)}}
\end{align*}](http://latex.artofproblemsolving.com/c/0/e/c0ebc3c6ea51d9627d8a8d1992cf5ad90b9408f9.png)
So, we want to show
Notice,
which is clearly true by AM-GM. Multiplying cyclically yields the desired result.
.