Solving a three variable equation I have three given values, suppose a=1.86, b=2.6 and c=4.2. Now I have to figure out x,y,z such that 


*

*$x\gt 0,y\gt 0$ and $z\gt 0$   

*$x+y+z=1$  

*$a*x\gt 1, b*y\gt 1$ and $cz\gt 1$  
I need a generalized solution steps for this to implement in programming.
Thanks.
 A: Since we need all of $ax, by, cz$ to be greater than $1$, given a set of $(a,b,c)$ we want $x,y,z$ to be bigger than $\frac{1}{a},\frac{1}{b},\frac{1}{c}$.
To find a generalized set of steps, first calculate $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. If this sum is greater than $1$, there are no solutions for $x,y,z$. This makes sense, because if we let $a,b,c$ to be $0.0001$ or some small number, $x,y,z$ would all have to be large and they would not sum to $1$, but we need $x+y+z=1$.
If it is less than or equal to $1$, then simply picking some number larger than $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ for $x,y,z$ will do.
If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<1$, we calculate $1-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$. Call this value $n$. Find the value of $\frac{n}{3}$ and add it to $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ and you will surely have values of $x,y,z$ that sum to $1$.
A: $\left\{ {\begin{array}{*{20}{l}}
{x > \frac{1}{a} \Rightarrow x = \frac{1}{a} + X}\\
{y > \frac{1}{b} \Rightarrow y = \frac{1}{b} + Y}\\
{z > \frac{1}{c} \Rightarrow z = \frac{1}{c} + Z}
\end{array}} \right.$  where $0 < X,Y,Z$
$   \Rightarrow \frac{1}{a} + X + \frac{1}{b} + Y + \frac{1}{c} + Z = 1 \Rightarrow X + Y + Z = 1 - \frac{1}{a} - \frac{1}{b} - \frac{1}{c}$ $\hspace{.3cm}$(*)
(*) is an equation of surface where cut each axis in $1 - \frac{1}{a} - \frac{1}{b} - \frac{1}{c}$. 
Your solution is all the point on this surface which that lie on the first quadrant.
A: $\left\{ \begin{array}{l}
x > \frac{1}{a} \Rightarrow x = \frac{1}{a} + \varepsilon \\
y > \frac{1}{b} \Rightarrow y = \frac{1}{b} + \varepsilon \\
z > \frac{1}{c} \Rightarrow z = \frac{1}{c} + \varepsilon 
\end{array} \right.$ where $0<\varepsilon$.
$\Rightarrow \frac{1}{a} + \varepsilon  + \frac{1}{b} + \varepsilon  + \frac{1}{c} + \varepsilon  = 1 \Rightarrow \varepsilon  = \frac{1}{3} - \frac{1}{{3a}} - \frac{1}{{3b}} - \frac{1}{{3c}} = {\rm{0}}{\rm{.1053}}$
