[/quote]
,
that satisfy the equation![\[
f(x) + 3f(-x) + f\left( \frac{1}{x} \right) = x, \quad \text{for all } x \in \mathbb{R}^*.
\]](http://latex.artofproblemsolving.com/6/a/2/6a2e35c42644036a62451b5e1794b36ffa728591.png)
Prove that
from positive integers to themselves st:
holds for any 
?
Subject to the constraints:
in primes 
.
angle 
is greater tha
.
is midpoint of 
.
and 
are the feet of the altitude from 
and 
are midpoint of 
and 
respectively. If 
intersect the line through 
parallel to 
,
.
,
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?
and circumcircle 
.
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.
?
Y by
Hello everyone,
I am learning functional equations.
To prove the below problem one -one function,I have taken two non-negative real numbers
from the domain 
and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!
I am learning functional equations.
To prove the below problem one -one function,I have taken two non-negative real numbers
from the domain 
and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!Attachments:
This post has been edited 3 times. Last edited by Vulch, Apr 11, 2025, 8:07 PM
distinct inputs, they cannot give the same output.
is a function. 
.
is one-one then for any choice of 
such that 
,
cannot be equal to 
,
.
and 
.
,
is not a correct example to contradict injectivity of the function.
such that 
.
.
,
.
be any nonzero real number, for surjectivity we need to show that there is a 
with 
.
,
and 
.
.
such that 
.
