Find all functions for $x+f(\frac1x) = 2f(x)$ This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with induction, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Find all functions $f:\mathbb{R}\setminus\{0\}\rightarrow \mathbb{R}$ such that, for all $x\in \mathbb{R}\setminus\{0\}$,
$$x+f\left(\frac1x\right) = 2f(x).$$

 A: Using Robert's comment,
If we put $y=\frac {1}{x} $ then
$$2f (x)-f (y)=x$$
$$2f (y )-f (x)=y $$
hence
$$3f (x )=2x+y $$
and
$$f (x)=\frac {1}{3}\left(2x+\frac {1}{x}\right)$$
A: You are given the equation
$$x+f\left(\frac1x\right) = 2f(x)$$
Following Robert Israel's hint, replace $\frac{1}{x}$ for $x$ and get
$$\frac{1}{x}+f(x) = 2f\left(\frac{1}{x}\right)$$
Combine the two equations to get
$$f(x) = \frac{2}{3}x+\frac{1}{3x}$$ 
A: Succumbing to my
inclination to generalize,
I'll look at
$x+af\left(\frac1x\right) 
= bf(x)
$.
Then
$f(\frac1{x})
=\frac{b}{a}f(x)-\frac{x}{a}
$.
Putting $\frac1{x}$ for $x$,
this becomes
$\frac1{x}+af(x) = bf(\frac1{x})
$
or
$\begin{array}\\
f(x)
&=\frac{b}{a}f(\frac1{x})-\frac{1}{ax}\\
&=\frac{b}{a}(\frac{b}{a}f(x)-\frac{x}{a})-\frac{1}{ax}\\
&=\frac{b^2}{a^2}f(x)-\frac{bx}{a^2}-\frac{1}{ax}\\
\text{or}\\
f(x)\frac{a^2-b^2}{a^2}
&=-\frac{bx}{a^2}-\frac{1}{ax}\\
\text{or}\\
f(x)(b^2-a^2)
&=bx+\frac{a}{x}\\
\text{or}\\
f(x)
&=\frac1{(b^2-a^2)}(bx+\frac{a}{x})\\
\end{array}
$
For $a=1, b=2$,
this gives
$f(x)
=\frac1{3}(2x+\frac{1}{x})
$.
Note that
there is no solution if
$a=b$.
This is because we get
$f(x)
=f(\frac1{x})+\frac{x}{a}
=f(x)+\frac{1}{xa}+\frac{x}{a}
$.
This method can
also solve the
apparently more general
$g(x)+af\left(\frac1x\right) 
= bf(x)
$.
The result will depend on
$g(x)$ and $g(\frac1{x})$.
