Find minimum of $\frac{a+3c}{a+2b+c}+\frac{7a+6b+3c}{a+b+2c}+\frac{c-a}{2a+b+c}$ for non-negative reals 
Let $a, b, c\ge 0$, not all zero. Find the minimum of
$$N = \frac{a+3c}{a+2b+c}+\frac{7a+6b+3c}{a+b+2c}+\frac{c-a}{2a+b+c}. $$

 A: (This initial part is not a solution. It is too long to be a comment. This demonstrates that the suggested approach of substituting the denominator doesn't work out because equality is not achieved as the condition $ a \geq 0$ is not met.)
We use the substitution $x = a + 2b + c, y = a+b+2c, z = 2a + b + c$.
This system gives us $ a = (-x-y+3z)/4, b = (3x-y-z)/4, c = (-x+3y-z)/4$.
The expression becomes:
$$\frac{ -x + 2y } { x} + \frac{ 2 x - y + 3z } { y} + \frac{ y - z  } { z} = \frac{ 2y}{x} + \frac{ 2x}{y} + \frac{3z}{y} + \frac{ y}{z}  -3. $$
Since $ \frac{ 2y}{x} + \frac{ 2x}{y} \geq 2\sqrt{4} = 4$ and $ \frac{ 3z}{y} + \frac{y}{z} \geq 2 \sqrt{3} $, hence a lower bound is $ 2 \sqrt{3} +1 \approx 4.46 $.
Equality is achieved when $ y = x,  y = \sqrt{3} z$, or equivalently when
$ a : b : c = 3 - 2 \sqrt{3} : 2 \sqrt{3} -1 : 2 \sqrt{3} - 1 $, which we have to verify have the same sign.
However, they do not have the same sign, so equality CANNOT be achieved. The minimum, or infimum, is higher.
We need the extra condition that (symmetric) $3z  \geq x + y$ in order to get (symmetric) $a \geq 0$.
This condition was violated when we tried to find the equality case.

Since the expression is symmetric, WLOG we may assume that $ y = 1$. If so, we want to minimize
$$ \frac{2}{x} + 2x + 3z + \frac{1}{z} - 3 $$
subject to $3z \geq x + 1, 3 \geq x + z, 3x \geq z + 1$.
Even doing Lagrangian, this is horrendous and has very ugly solutions. We can show that the minimum happens on the boundary $3z=x+1$, but the actual solution ($\approx 4.493$) is pretty ugly.
I'm doubtful there's a contest-math solution.
My guess is that the problem setters made a mistake by not checking $ a : b : c$.
I do wish there was a nice way to complete this problem though, since it would illustrate the importance of checking conditions instead of just assuming as most of us did.
A: Hint: Let $x,y,z$ be the denominators of the three fractions, and express $a,b,c$ in terms of $x,y,z$. Then write $N$ in terms of $x,y,z$. After some simplification we will get terms like $\frac{x}{z},\frac{x}{y},\frac{y}{x},\frac{y}{z},\cdots$Then consider inequality $\frac{x}{y}+\frac{y}{x}\geqslant2$ if $xy$ is positive and analogously $\frac{x}{z}+\frac{z}{x}\geqslant2$ and so on...
