WLOG, let $x+y+z=1$ so $\frac{|x|+|x+4y|+|y+7z|+2|z|}{|x+y+z|}=|x|+2|z|+|4-3x-4z|+|1-x+6z|$ suppose $f(z)=|x|+2|z|+|4-3x-4z|+|1-x+6z|$ it's trival that $f(z) \geq f(\dfrac {x-1} 6)$ so $|x|+2|z|+|4-3x-4z|+|1-x+6z| \geq |x|+\dfrac {|x-1|} 3 +|\dfrac {11x} 3-\dfrac {14} 3|$ suppose $g(x)=|x|+\dfrac {|x-1|} 3 +|\dfrac {11x} 3-\dfrac {14} 3|$, it' also trival that $g(x) \geq g(\dfrac {14} {11})$ so the minimum value is $\dfrac {15} {11}$, the equation holds when $x=-4y=28z$

Posted previously by me: https://artofproblemsolving.com/community/u488422h2923667p26140565