This one impossible

I concur.

I somehow "proved" that $f(x)=1/3$ is the only solution kms.

Uh got all of these things: f(2x+1) = 1/2 (1 + f(x)) - f(1) f(x) + f(1/x) = 2 f(1) 2 f(x+1) + f(2/x) = 1 f(2x+1) + f(1 + [2/x]) = 1 - f(1) Got solution set as $\frac{3-6c}{3} + \frac{6c-2}{x+1}$ where $\frac{1}{3} \le c \le \frac{1}{2}$.

EDITED 2018-04-23: some cleanup and filling in more details. As always, let $x=b/c$, $y=c/a$, $z=a/b$ (classical inequality trick). Then the problem becomes \[ \sum_{\text{cyc}} f\left( \frac{b+c}{a} \right) = 1. \]Let $f(t) = g(\frac{1}{t+1})$, equivalently $g(s) = f(1/s-1)$. Thus $g \colon (0,1) \to (0,1)$ which satisfies $\sum_{\text{cyc}} g\left( \frac{a}{a+b+c} \right) = 1$, or equivalently \[ \boxed{g(a) + g(b) + g(c) = 1} \qquad \forall a+b+c=1. \] The rest of the solution is dedicated to solving this equivalent functional equation in $g$. It is a lot of technical details and I will only outline them (with apologies to the contestants who didn't have that luxury). Claim: The function $g$ is linear. Proof. This takes several steps, all of which are technical. We begin by proving $g$ is linear over $(1/8, 3/8)$. First, whenever $a+b \le 1$ we have \[ 1 - g(1-(a+b)) = g(a) + g(b) = 2 g\left( \frac{a+b}{2} \right). \]Hence $g$ obeys Jensen's functional equation over $(0,1/2)$. Define $h \colon [0,1] \to {\mathbb R}$ by $h(t) = g(\frac{2t+1}{8}) - (1-t) \cdot g(1/8) - t \cdot g(3/8)$, then $h$ satisfies Jensen's functional equation too over $[0,1]$. We have also arranged that $h(0) = h(1) = 0$, hence $h(1/2) = 0$ as well. Since \[ h(t) = h(t) + h(1/2) = 2h(t/2+1/4) = h(t+1/2) + h(0) = h(t+1/2) \]for any $t < 1/2$, we find $h$ is periodic modulo $1/2$. It follows one can extend $\widetilde h$ by \[ \widetilde h \colon {\mathbb R} \to {\mathbb R} \qquad\text{by}\qquad \widetilde h(t) = h(t - \left\lfloor t \right\rfloor) \]and still satisfy Jensen's functional equation. Because $\widetilde h(0) = 0$, it's well-known this implies $\widetilde h$ is additive (because $\widetilde h(x+y) = 2\widetilde h\left( (x+y)/2 \right) = \widetilde h(x) + \widetilde h(y)$ for any real numbers $x$ to $y$). But $\widetilde h$ is bounded below on $[0,1]$ since $g \ge 0$, and since $\widetilde h$ is also additive, it follows (well-known) that $\widetilde h$ is linear. Hence so is $h$. So, the function $g$ is linear over $(1/8,3/8)$; thus we may write $g(x) = kx + \ell$, valid for $1/8 < x < 3/8$. Since $3g(1/3) = 1$, it follows $k + 3\ell = 1$. For $0 < x < 1/8$ we have $g(x) = 2g(0.15) - g(0.3-x) = 2(0.15k+\ell) - (k(0.3-x)+\ell) = kx+\ell$, so $g$ is linear over $(0,3/8)$ as well. Finally, for $3/8 < x < 1$, we use the given equation \[ 1 = g\left( \frac{1-x}{2} \right) + g\left( \frac{1-x}{2} \right) + g(x) \implies g(x) = 1 - 2\left( k \cdot \frac{1-x}{2} + \ell \right) = kx+\ell \]since $\frac{1-x}{2} < \frac{5}{16} < \frac{3}{8}$. Thus $g$ is linear over all. $\blacksquare$ Putting this back in, we deduce that $g(x) = kx + \frac{1-k}{3}$ for some $k \in [-1/2,1]$, and so \[ f(x) = \frac{k}{x+1} + \frac{1-k}{3} \]for some $k \in [-1/2,1]$. All such functions work.

pretty sure k is from -1/2 to 1

wow from reading these solutions, this is so much more impossible than i thought

lucasxia01 wrote:

ah you got me there

v_Enhance wrote: As always, let $x=b/c$, $y=c/a$, $z=a/b$ (classical inequality trick). Then the problem becomes \[ \sum_{\text{cyc}} f\left( \frac{b+c}{a} \right) = 1. \]Let $f(t) = g(\frac{1}{t+1})$, equivalently $g(s) = f(1-1/s)$. Thus $g$ is a function $(0,1) \to (0,1)$ which satisfies $\sum_{\text{cyc}} g\left( \frac{a}{a+b+c} \right) = 1$, or equivalently \[ \boxed{g(a) + g(b) + g(c) = 1} \qquad \forall a+b+c=1. \] The rest of the solution is dedicated to solving this equivalent functional equation in $g$. It is a lot of technical details and I will only outline them (with apologies to the contestants who didn't have that luxury). Claim: The function $g$ is linear over $(1/3,2/3)$. Proof. This takes several steps, all of which are technical. First, whenever $a+b \le 1$ we have \[ 1 - g(1-(a+b)) = g(a) + g(b) = 2 g\left( \frac{a+b}{2} \right). \]Hence $g$ obeys Jensen's functional equation over $(0,1/2)$. Define $h \colon [0,1] \to {\mathbb R}$ by $h(t) = g(t/3+1/3) - (1-t)g(1/3) - tg(2/3)$, then $h$ satisfies Jensen's functional equation too. We have also arranged that $h(0) = h(1) = 0$, hence $h(1/2) = 0$ as well. Since \[ h(t) = h(t) + h(1/2) = 2h(t/2+1/4) = h(t+1/2) + h(0) = h(t+1/2) \]for any $t < 1/2$ we find $h$ is periodic with period $1/2$. From this, one can extend $\tilde h$ by \[ \tilde h \colon {\mathbb R} \to {\mathbb R} \qquad\text{by}\qquad \tilde h(t) = h(t - \left\lfloor t \right\rfloor) \]and still satisfy Jensen's functional equation. Hence $\tilde h$ is additive. But $\tilde h$ is bounded below on $[0,1]$ since $g \ge 0$, so $\tilde h$ is linear, thus so is $h$, as desired. $\blacksquare$ From this, one can quickly show $g$ is linear over $(0,1)$. Thus, $g(x) = kx + \frac{1-k}{3}$ for some $k \in [0,1]$, and so \[ f(x) = \frac{k}{t+1} + \frac{1-k}{3} \]for some $k \in [0,1]$. All such functions work. I did a similar thing though after I got Jensen's I proved that it was linear over the rationals and proved continuity. Also, I wrote my solution set as $k(\frac{1}{x+1}-\frac{1}{3})+\frac{1}{3}$. Like r3mark, I got $k \in [-\frac{1}{2}, 1]$

I couldn't even touch this problem

Do I get a point for finding the correct form and proving that it works?

Superwiz wrote: Uh got all of these things: f(2x+1) = 1/2 (1 + f(x)) - f(1) f(x) + f(1/x) = 2 f(1) 2 f(x+1) + f(2/x) = 1 f(2x+1) + f(1 + [2/x]) = 1 - f(1) Got solution set as $1-2c + \frac{6c-2}{x+1}$ where $\frac{1}{3} \le c \le \frac{1}{2}$. Do you think I can get partials for any of this?

vitriolhumor wrote: wow from reading these solutions, this is so much more impossible than i thought I think it's long and technically involved, but there is no idea, other than the substitution at the beginning. It's unclear how much of the latter stuff can be cited, too (you could shorten my solution a lot by simply claiming Jensen's f.e. over an interval extends to an additive function, which would make the solution super terse).

Let $g(x)=f\left(\frac{1}{x}-1\right)$ ($g$ has domain $(0,1)$ and obviously range bounded by $(0,1)$). Then if $a+b+c=1$ \[g(a)+g(b)+g(c)=f(\frac{1}{a}-1)+f(\frac{1}{b}-1)+f(\frac{1}{c}-1)=\sum_{cyc}f(\frac{a}{b}+\frac{1}{\frac{b}{c}})=1.\]Now let $h(x)=g(x+\frac{1}{3})-\frac{1}{3}$, with domain $(-\frac{1}{3},\frac{2}{3})$ and range bounded by $(-\frac{1}{3},\frac{2}{3})$. $h$ satisfies when $x+y+z=0$ we have $h(x)+h(y)+h(z)=0$. Letting $z=0$ we have $h$ is odd on $(-\frac{1}{3},\frac{1}{3})$, then plugging $z=-x-y$ for $x,y\in(-\frac{1}{6},\frac{1}{6})$ we get $h(x+y)=h(x)+h(y)$, and using $2h(x)+h(-2x)=0$ we get that $h(x+y)=h(x)+h(y)$ on the whole interval. But this is Cauchy's functional equation (the proof for the real line works verbatim for the interval), and $h$ is bounded on the interval, so $h(x)=cx$ for some $c\in\mathbb{R}$. Back substituting we get $f(x)=\frac{1-c}{3}+\frac{c}{x+1}$, and it's easy to check this works iff $-\frac{1}{2}\le c\le 1$.

v_Enhance wrote: As always, let $x=b/c$, $y=c/a$, $z=a/b$ (classical inequality trick). Then the problem becomes \[ \sum_{\text{cyc}} f\left( \frac{b+c}{a} \right) = 1. \]Let $f(t) = g(\frac{1}{t+1})$, equivalently $g(s) = f(1-1/s)$. Thus $g$ is a function $(0,1) \to (0,1)$ which satisfies $\sum_{\text{cyc}} g\left( \frac{a}{a+b+c} \right) = 1$, or equivalently \[ \boxed{g(a) + g(b) + g(c) = 1} \qquad \forall a+b+c=1. \] The rest of the solution is dedicated to solving this equivalent functional equation in $g$. It is a lot of technical details and I will only outline them (with apologies to the contestants who didn't have that luxury). Claim: The function $g$ is linear over $(1/3,2/3)$. Proof. This takes several steps, all of which are technical. First, whenever $a+b \le 1$ we have \[ 1 - g(1-(a+b)) = g(a) + g(b) = 2 g\left( \frac{a+b}{2} \right). \]Hence $g$ obeys Jensen's functional equation over $(0,1/2)$. Define $h \colon [0,1] \to {\mathbb R}$ by $h(t) = g(t/3+1/3) - (1-t)g(1/3) - tg(2/3)$, then $h$ satisfies Jensen's functional equation too. We have also arranged that $h(0) = h(1) = 0$, hence $h(1/2) = 0$ as well. Since \[ h(t) = h(t) + h(1/2) = 2h(t/2+1/4) = h(t+1/2) + h(0) = h(t+1/2) \]for any $t < 1/2$ we find $h$ is periodic with period $1/2$. From this, one can extend $\tilde h$ by \[ \tilde h \colon {\mathbb R} \to {\mathbb R} \qquad\text{by}\qquad \tilde h(t) = h(t - \left\lfloor t \right\rfloor) \]and still satisfy Jensen's functional equation. Hence $\tilde h$ is additive. But $\tilde h$ is bounded below on $[0,1]$ since $g \ge 0$, so $\tilde h$ is linear, thus so is $h$, as desired. $\blacksquare$ From this, one can quickly show $g$ is linear over $(0,1)$. Thus, $g(x) = kx + \frac{1-k}{3}$ for some $k \in [-1/2,1]$, and so \[ f(x) = \frac{k}{t+1} + \frac{1-k}{3} \]for some $k \in [-1/2,1]$. All such functions work. this is an amazing solution how long did you take to come up with this

The solution is pretty intuitive (but bashy) once you get to the point where $g(a)+g(b)+g(c) = 1$. I think getting to that part (realizing that $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 1$ where $a = x+\frac{1}{y}$ and $b, c$ are defined similarly) is the hard part. My solution was nine pages for reference, though I tried to make it completely airtight.

how many points would I get if I got $g(a)+g(b)+g(c)=1$ and the correct final solution set but didn't have much of a proof? I cited cauchy's fe instead of jensen oops

lucasxia01 wrote:

Jenson > Mt. Inequality Eruption

Can someone please check this?

Above I don't think this works as it doesn't follow the second case or maybe it does and I'm too dumb to see it

TheBigOne wrote: I somehow "proved" that $f(x)=1/3$ is the only solution kms. 2018 humor be like

zephy wrote: Above I don't think this works as it doesn't follow the second case or maybe it does and I'm too dumb to see it What do you mean by "not following the second case"? If you are referring to the $t,y$ equations, then those are symmetric to the $x,z$ equations, and so they hold. (By the way, I got @CyclicISLscelesTrapezoid to proofread it and he said it was fine.)

Math4Life2020 wrote: zephy wrote: Above I don't think this works as it doesn't follow the second case or maybe it does and I'm too dumb to see it What do you mean by "not following the second case"? If you are referring to the $t,y$ equations, then those are symmetric to the $x,z$ equations, and so they hold. (By the way, I got @CyclicISLscelesTrapezoid to proofread it and he said it was fine.) Oops my bad I didn't see the symmetry

Math4Life2020 wrote: Can someone please check this? Seems OK to me.

I think it is impossible for me to solve it.

Math4Life2020 wrote:

Can someone please check this? I am confused - doesn't $x+y=z+t$ directly imply $g(x)+g(y)=1-g(1-x-y)=1-g(1-z-t)=g(z)+g(t)$? am I missing something? @below I see, thanks!

polarbear08 wrote: Math4Life2020 wrote:

Can someone please check this? I am confused - doesn't $x+y=z+t$ directly imply $g(x)+g(y)=1-g(1-x-y)=1-g(1-z-t)=g(z)+g(t)$? am I missing something? x+y could be, say, 4/3.

Here is a writeup from the corresponding OTIS walkthrough for details practice. The answer is $f(x) = \frac k{x+1} + \frac{1-k}3$ for any $-\frac 12 \leq k \leq 1$. In particular, setting $x= a/b$ and cyclic variants yields $$f\left(\frac{a+c}b\right) + f\left(\frac{b+a}c\right) + f\left(\frac{c+b}a\right) = 1.$$We may set $g(1/(x+1)) = f(x)$ and de-homogenize to get a function $g':(0, 1) \to (0, 1)$ with $$g(a')+g(b')+g(c') = 1$$for all $a'+b'+c' = 1$. The idea now is to turn this into an additive plus bounded equals linear setup. First, note that because $$g(a') + g(b') = 1-g(1-a'-b') = 2g\left(\frac{a'+b'}2\right)$$by applying the conclusion twice, $f$ is Jensen for all $a' + b' \leq 1$, in particular on $\left(0, \frac 12\right)$. So then we ought to define a function $h$ such that $h(0) = h(1) = 0$ that is analogous to $g$ on some nontrivial interval; this $h$ is given explicitly by $$h(x) = g\left(\frac{2x+1}8\right) - (1-x)g\left(\frac 18\right) - xg\left(\frac 38 \right).$$In particular, $g = h$ on $\left[\frac 18, \frac 38\right]$, and so $h$ is also Jensen over $[0, 1]$. We may extend $h$ to $\mathbb R$ by noting that $$h\left(x+\frac 12\right) = h(x) + h\left(\frac 12\right) = h(x)$$where $h$ Jensen implies additive. But because $h$ is bounded below on $[0, 1]$, we may finally conclude that $h$ is linear and correspondingly $h \equiv 0$. Thus $g$ is linear; set $g(x) = kx + \ell$. By $3g\left(\frac 13\right) = 1$ follows $k+3\ell = 1$. Now we can extend $g$ to $\left(0, \frac 18\right)$ by writing $$g(\varepsilon) + g\left(\frac 38\right) = 2g\left(\frac 18 < c < \frac 38\right)$$and similarly $$g(1-\varepsilon) + g(\varepsilon_1) + g(\varepsilon_2) = 1$$extends $g$ to $\left(\frac 38, 1\right)$ too. Finally, as $k + \ell \geq 0$ and $\ell \geq 0$ by considering $g(0)$ and $g(1)$, it follows that $-\frac 12 \leq k \leq 1$, which yields the desired class of solutions.

HamstPan38825 wrote: Here is a writeup from the corresponding OTIS walkthrough for details practice. The answer is $f(x) = \frac k{x+1} + \frac{1-k}3$ for any $-\frac 12 \leq k \leq 1$. In particular, setting $x= a/b$ and cyclic variants yields $$f\left(\frac{a+c}b\right) + f\left(\frac{b+a}c\right) + f\left(\frac{c+b}a\right) = 1.$$We may set $g(1/(x+1)) = f(x)$ and de-homogenize to get a function $g':(0, 1) \to (0, 1)$ with $$g(a')+g(b')+g(c') = 1$$for all $a'+b'+c' = 1$. The idea now is to turn this into an additive plus bounded equals linear setup. First, note that because $$g(a') + g(b') = 1-g(1-a'-b') = 2g\left(\frac{a'+b'}2\right)$$by applying the conclusion twice, $f$ is Jensen for all $a' + b' \leq 1$, in particular on $\left(0, \frac 12\right)$. So then we ought to define a function $h$ such that $h(0) = h(1) = 0$ that is analogous to $g$ on some nontrivial interval; this $h$ is given explicitly by $$h(x) = g\left(\frac{2x+1}8\right) - (1-x)g\left(\frac 18\right) - xg\left(\frac 38 \right).$$In particular, $g = h$ on $\left[\frac 18, \frac 38\right]$, and so $h$ is also Jensen over $[0, 1]$. We may extend $h$ to $\mathbb R$ by noting that $$h\left(x+\frac 12\right) = h(x) + h\left(\frac 12\right) = h(x)$$where $h$ Jensen implies additive. But because $h$ is bounded below on $[0, 1]$, we may finally conclude that $h$ is linear and correspondingly $h \equiv 0$. Thus $g$ is linear; set $g(x) = kx + \ell$. By $3g\left(\frac 13\right) = 1$ follows $k+3\ell = 1$. Now we can extend $g$ to $\left(0, \frac 18\right)$ by writing $$g(\varepsilon) + g\left(\frac 38\right) = 2g\left(\frac 18 < c < \frac 38\right)$$and similarly $$g(1-\varepsilon) + g(\varepsilon_1) + g(\varepsilon_2) = 1$$extends $g$ to $\left(\frac 38, 1\right)$ too. Finally, as $k + \ell \geq 0$ and $\ell \geq 0$ by considering $g(0)$ and $g(1)$, it follows that $-\frac 12 \leq k \leq 1$, which yields the desired class of solutions. bump, but how would one think to define such a function $h$?

We claim that the answer is $$f(x)=\frac{c}{x+1}+\frac{1}{3}-\frac{1}{3}c$$for $-\frac{1}{2}\leq c\leq 1$. This works since it is a weighted average of $\frac{1}{3}$ and $\frac{1}{x+1}$ which both work. First, substitute $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$ for $a,b,c>0$. The equation becomes $$\sum_{cyc}f(\frac{b+c}{a})=1.$$ Now, since this is homogenous, assert $a+b+c=1$ so that it's $$\sum_{cyc}f(\frac{1}{a}-1)=1.$$ Now, let $g(x)=f(\frac{1}{x}-1),$ so that $$g(a)+g(b)+g(c)=1$$whenever $a+b+c=1$. Finally, let $g(x)=\frac{1}{3}+h(x-\frac{1}{3})$ so that $$h(x)+h(y)+h(z)=0$$whenever $x+y+z=0$. Note that $h(x)>\frac13$ as $g(x)>0$. We claim that $h(x)=cx$. First, $(x,y,z)=(0,x,-x)$ gives $h(x)=-h(-x)$ for $-\frac13<x<\frac13$. Then, $(x,y,z)=(x,x,-2x)$ gives $2h(x)=-h(-2x)$ for $-\frac13<x<\frac16$. Combining with the previous, $2h(x)=h(2x)$ for $-\frac13<x<\frac16$. $z=-x-y$ gives $h(x)+h(y)=-h(-x-y)$. As long as $\frac{-x-y}{2}$ is in $(-1/3,1/6)$, that is, $x+y$ is in $(-1/3,2/3)$, we have that $$-h(-x-y)=-2h(\frac{-x-y}{2})$$, after which we can apply the odd claim (as it is valid between -1/3 and 1/3) to get $-2h((-x-y)/2)=2h((x+y)/2$. Thus, for $-\frac13<x+y<2/3$, we have that $h(x)+h(y)=2h((x+y)/2)$. Consider the region $R$ of the $xy$ plane between $x=-\frac16$ and $x=\frac13$. Split this into $5$ regions, which we call $A$, $B$, $C$, $D$, and $Dead$, as shown in the diagram attached Since $h(x)>-\frac13$, $h$ should not contain any points in $Dead$. (the boundaries are considered to be in none of the regions). The condition $$h(x)+h(y)=2h(\frac{x+y}{2})$$tells us two things: given any two points in $R$ that are on $h$, their midpoint also is, and that the reflection of one of the points over the other is on $h$ as long as it lies in $R$. Due to this, given any line $\ell$ containing two points of $h$ in $R$, the set of points on $\ell$ that are also on $h$ is dense on the region of $\ell$ inside $R$, as we can reach any point whose weight has a denominator a power of $2$. A corrolary of this is that, if such a line $\ell$ passes through the interior of $Dead$, then $h$ contains points of $Dead$ as it is dense. Thus, this cannot happen. If a point $P$ is in the interior of region $A$ or $D$, then the line through it and $(0,0)$ will pass through $Dead$, contradiction. Thus, $h$ also contains no points of $A$ or $D$, so $A$ and $D$ now functions the same as $Dead$ as no points of $h$ can be in it. The only way any non-vertical line can avoid all of $A$, $D$, and $Dead$ is by passing through $(0,0)$. Thus, $h$ is linear, as desired. Reversing transformations recovers $$f(x)=\frac{c}{x+1}+\frac{1}{3}-\frac{1}{3}c.$$ However, we must have $-\frac{1}{2}\leq c\leq 1$ to ensure that $f$ stays within the allowed range.

The solution set is all functions $f$ of the form \[C \left( \frac{1}{x+1} - \frac{1}{3} \right) - \frac{1}{3}\]for $- \tfrac{1}{2} < C < 1$. Let $x = \tfrac{a}{b}$, $y = \tfrac{b}{c}$ and $z = \tfrac{c}{a}$. Let $g : (0, 1) \rightarrow (0, \infty)$ be the function such that \[g(x) = f \left( \frac{1}{x} - 1 \right) \qquad \Leftrightarrow \qquad g\left( \frac{1}{x+1} \right) = f(x).\]With all these substitutions, our functional equation is equivalent to solving \[g(a)+g(b)+g(c) = 1\]when $a+b+c = 1$. This equation reduces to Cauchy's as follows: first, if $a+b=1$, we have \begin{align*} & g(a)+g(b) \\ = ~ & 2 - 2 \left( g \left( \tfrac{1-a}{2} \right) + g \left( \tfrac{1-b}{2} \right) \right) \\ = ~ & 2 - 2 \left( 1 - g( \tfrac{1}{2} ) \right) \\ = ~ & 2 g ( \tfrac{1}{2} ). \end{align*}Now, we have \[ g(a) + g(b) = 1 - g(1-a-b) = g(a+b) + 1 - 2 g ( \tfrac{1}{2} ), \]so $g(x) + \tfrac{1}{2} - g( \tfrac{1}{2})$ satisfies Cauchy's equation over the interval $( 0, \tfrac{1}{2} )$. However, since this function is bounded from below, it must be linear, so $g(x)$ is linear. This gives us all the solutions we initially claimed.

Solved with iamhungry. We claim that the only solutions are all functions $f$ in the form of \[f(x)=p\left(\frac{x}{x+1}\right)+\frac{1-2p}{3},\]where $p$ is any real number in the interval $\left[-1,\frac{1}{2}\right]$. *** First, we begin with the following lemma. *** Lemma. If there are real numbers $a$, $b$, $c$ such that \[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1,\]then there exist $x$, $y$, $z$ such that $a=x+\frac{1}{y}$, $b=y+\frac{1}{z}$, and $c=z+\frac{1}{x}$ and $xyz=1$. Additionally, if $xyz=1$, then $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1$, where $a=x+\frac{1}{y}$, $b=y+\frac{1}{z}$, and $c=z+\frac{1}{x}$. Proof. We will first prove the latter half of the claim. Note that $z=\frac{1}{xy}$. Then, \[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1},\]\[=\frac{1}{1+x+\frac{1}{y}}+\frac{1}{y+xy+1}+\frac{1}{1+\frac{1}{x}+\frac{1}{xy}},\]\[=\frac{y}{y+xy+1}+\frac{1}{y+xy+1}+\frac{xy}{y+xy+1}=1,\]as desired. The fact that there always exists $x$, $y$, $z$ can be proved by solving the system of equations \[a=x+\frac{1}{y},\]\[b=y+\frac{1}{xy},\]and the value of $c$ that works is then forced by the latter half of the claim, as desired. *** Now, define a function $g$ with domain of $(-\infty,\infty)\backslash\{0,1\}$ so that for $x\in (0,\infty)$, $f(x)=g\left(1-\frac{1}{1+x}\right)$. Note that $g$ is bounded over the interval $(0,1)$ to be in the range $(0,1)$ because these are the values of $g$ that yield values yielded by $f$. From the problem constraints, we can conclude that all values of $f$ are in the range $(0,1)$, meaning that $g$ is bounded in $(0,1)$ over $(0,1)$. We will use this fact later. For now, again let $a=x+\frac{1}{y}$, $b=y+\frac{1}{z}$, and $c=z+\frac{1}{x}$ for $xyz=1$. Then, by problem conditions, we have that \[f(a)+f(b)+f(c)=1,\]\[\implies g\left(1-\frac{1}{a+1}\right)+g\left(1-\frac{1}{b+1}\right)+g\left(1-\frac{1}{c+1}\right)=1,\]and since $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1$, we can let $m=1-\frac{1}{1+a}$ and $n=1-\frac{1}{1+b}$ to get \[g(m)+g(n)+g(2-m-n)=1,\]and by The Lemma, since there are always values of $x$, $y$, $z$ that get the values of $a$, $b$, $c$ to show up, this holds for all $m$, $n$. Then, define $g(2)$ to be a constant, say $d$. Setting $n=-m$, we get \[g(x)+g(-x)=1-g(2)=1-d,\]so $g(-x)=1-d-g(x)$. Then, let us consider our equation: \[g(m)+g(n)+g(2-m-n)=1,\]\[\implies g(m)+g(n)=1-g(2-m-n)=1-(-g(m+n-2)+1-d)=g(m+n-2)+d,\]since $g(-x)=1-d-g(x)$. Then, letting $h(x)=g(x+2)+d$ then gives \[h(m-2)+h(n-2)=h(m+n-4) \implies h(x)+h(y)=h(x+y),\]which is Cauchy's Functional Equation! We can apply Cauchy's since $g(x)$ is bounded over $(0,1)$, meaning that $h(x)$ must be bounded over $(0,1)$, which is sufficient for Cauchy's, as the condition is simply "bounded over any interval." This means that $h$ is linear, which in turn implies that $g$ is linear from working backwards. Now let $g(x)=px+q$. However, note that \[g(x)=g\left(1-\frac{1}{1+\left(-1+\frac{1}{1-x}\right)}\right),\]and since $g\left(1-\frac{1}{1+x}\right)=f(x)$, this means that \[\implies g(x)=f\left(-1+\frac{1}{1-x}\right)=f\left(\frac{x}{1-x}\right).\]Then, our problem becomes classifying the possibilities for $p$ and $q$. However, note that \[g(m)+g(n)+g(2-m-n)=1,\]\[\implies p(m+n+(2-m-n))+3q=1,\]\[\implies 2p+3q=1 \implies q=\frac{1-2p}{3},\]as the pairs of $p$ and $q$ that work here. Plugging this in gives that \[f\left(\frac{x}{1-x}\right)=g(x)=px+\frac{1-2p}{3}.\]However, note that \[f(x)=f\left(\frac{\frac{x}{x+1}}{1-\frac{x}{x+1}}\right)=p\left(\frac{x}{x+1}\right)+\frac{1-2p}{3},\]making this the general form for all possible solutions to $f(x)$. Experimenting with values of $p$ gives that the only values of $p$ that give us a function in our desired range are the ones in the range $\left[-1,\frac{1}{2}\right]$, as desired. There is no pointwise trap since we used Cauchy's FE to come up with $h$, which gave us a unique equation for $g$, which then in turn gave us a unique equation for $f$. We can then plug in our solution we've found into the problem conditions to check that it works. Again, we use $a$, $b$, $c$ for simplicity, to get that \[f(a)+f(b)+f(c),\]\[=p\left(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\right)+3\left(\frac{1-2p}{3}\right),\]however, note that since $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1$, $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$ is just $3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)$, or $2$, and plugging this in gives \[2p+3\left(\frac{1-2p}{3}\right)=2p+(1-2p)=1,\]as desired. Therefore our solution set does indeed work, and these are all of the solutions. This is what we wished to find, concluding our proof.