TITLE: What are $x$, $y$ and $z$ if $\frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y} = 4$ and $x$, $y$ and $z$ are whole numbers?
QUESTION [8 upvotes]: What are $x$, $y$ and $z$ if $$\dfrac{x}{y + z} + \dfrac{y}{x + z} + \dfrac{z}{x + y} = 4$$ and $x$, $y$ and $z$ are whole numbers?
MY ATTEMPT
Let $u = x + y + z$.  Then the equation can be rewritten as
$$\dfrac{x}{u - x} + \dfrac{y}{u - y} + \dfrac{z}{u - z} = 4$$
Suppose I set
$$1 = \dfrac{x}{u - x} = \dfrac{y}{u - y}$$
and
$$2 = \dfrac{z}{u - z}.$$
Then I get
$$x = y + z$$
$$y = x + z$$
$$z = 2(x + y),$$
so that
$$z = 0 = x + y,$$
which is impossible.
Next, suppose I set
$$\dfrac{4}{3} = \dfrac{x}{u - x} = \dfrac{y}{u - y} = \dfrac{z}{u - z}.$$
Then I get
$$4(u - x) = 3x$$
$$4(u - y) = 3y$$
$$4(u - z) = 3z$$
so that
$$12u - 4(x + y + z) = 3(x + y + z)$$
which implies that
$$12u = 7(x + y + z) = 7u$$
from which it follows that
$$u = 0.$$
This is, again, impossible.
Alas, here is where I get stuck.  Any hint(s) will be appreciated.

REPLY [7 votes]: If we set $$x+y=u, y+z= v, z+x=w  $$ 
We have $$x=\frac{u+w-v}{2}, y=\frac{u+v-w}{2}, z=\frac{w+v-u}{2}$$
Now $$2\sum_{cyc} \frac{z}{x+y}=\sum_{cyc}\frac{u+v-w}{w}=8 \Rightarrow \sum_{cyc} \frac{u+v}{w}=11$$
However, one can prove that the only naturals that can be expressed of the $$\sum_{cyc} \frac{a+b}{c}$$
Where $a,b,c \in \mathbb{N}$ are $6,7,8$. This is relatively simple, as we can assume $\text{WLOG}$ $$
a \ge b \ge c, \dfrac{b+c}{a} \in \mathbb{N}$$so I leave this to you. 
EDIT
The previous answer here is, unfortunately, incorrect. This is because I got confused and accidentally assumed that $(x,y,z)$ are all relatively coprime. However, as this is not implied in the question, my answer written here is incorrect. 
The correct answer is written here, where the user @Next gives us a solution for $$\sum_{cyc} \frac{a}{b+c}=4$$ It is the following. 

$a=$2332797891204725453580403814216955612718693675609518139813675622446336
  8530351921955206357565424226029748329737767516130520072674084336131550
  2597616224970927979227396663481447506019173462295157784788781420305046
  5201815993661680059006448575315523206103260762210944137954571975497854
  9786027663601160534574317253280344812956727894696796553762212813889660
  9065956718516224446015577143267128739011935697434909021669583635379832
  35022557869209259
$b=$1161640217306132458900911441651415023972393417197892812143262449233898
  8034221463466278254018560734492913221738943224762433374574861704275058
  0062902808034990817009121975186967451351814311101112040391014295321972
  8784138582766210837461563508481437266175417187186208008663435889653439
  7066554486263784443013141020886435995672339322997499528376940620045001
  1919735272479457688230567501843839892799164246003766614214017398378635
  0444307965016411
$c=$5054729227475450427274369484803239479825091305751388135572448603576037
  6549781961422098862259430557133842304461180359698183208339647924784255
  6816542651386138853492649101592171641096957016404851774814750638840260
  3496289958758089911825477669004739864966841494437579004665357462952425
  4130327474390635537868978719887059697148297723373566417781389238382736
  3204638301684342182024187145267526992579708085994452308601529371953916
  7125415529515145

And there are infinitely many other solutions. I apologize for my previously incorrect answer. There does exist such $x,y,z$.