Find minimum of ${(a+3c)\over(a+2b+c)} + {4b\over(a+b+2c)} - {8c\over(a+b+3c)}$ for non-negative reals 
Let $a, b, c \ge 0$, not all zero. Find the minimum value of $${(a+3c)\over(a+2b+c)} + {4b\over(a+b+2c)} - {8c\over(a+b+3c)}.$$

Here was my attempt:
If c tends to infinity and a and b are small we get 1/3.
Now , i took $b=0$ and we get $(a-c)^2\over(a+c)(a+3c)$ clearly at $a=c$ this is $0$. And hence i can conclude that the infimum is $0$. This looked optimal.
But I took $a=1, b=3, c=10$ where the result is negative.
I don't see any definitve approach to this problem now. Hence, requesting your help !

As Albus mentioned, we use the substitution:
$ x = a + 2b + c, y = a + b + 2c, z = a + b + 3c$.
This system gives us $ a = -x  + 5y - 3z, b = x - 2y + z, c = 0x - y + z$.
The expression becomes:
$$\frac{ -x + 2y } { x} + \frac{ 4 x - 8 y + 4z } { y} + \frac{ 8y - 8z  } { z} = \frac{ 2y}{x} + \frac{ 4x}{y} + \frac{4z}{y} + \frac{ 8y}{z}  -17. $$
Since $ \frac{ 2y}{x} + \frac{ 4x}{y} \geq 2\sqrt{8} = 4 \sqrt{2}$ and $ \frac{ 4z}{y} + \frac{8y}{z} \geq 2 \sqrt{32} = 8 \sqrt{2}$, hence a lower bound is $ 12 \sqrt{2} - 17 \approx -0.0294$.
Equality is achieved when $ y = \sqrt{2} x,  z = \sqrt{2} y$, or equivalently when
$ a : b : c = 5\sqrt{2} - 7 : 3 - 2 \sqrt{2} : 2 - \sqrt{2} $, which we have to verify have the same sign (If not, that can't be achieved.)
