Minimum value when $abc+ab+4bc+9ca=144$ If $a,b,c$ are non-negative real numbers such that $abc+ab+4bc+9ca=144$, find the minimum value of $a+b+c$.
I tried with Lagrange multipliers. I got the system:
$bc+b+9c=ca+a+4c=ab+4b+9a$
Replacing in the condition, I found four solutions, but only one $(4,0,4)$ is non-negative. So the minimum value is $8$. My question is, can this be done without Lagrange Multipliers?

Let $a+b+c<8,$ $a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $x+y+z=8.$
Thus, $$k(x+y+z)<8,$$ which gives $0<k<1.$
Thus, $$144=k^3xyz+k^2(xy+4yz+9zx)<xyz+xy+4yz+9zx,$$ which is a contradiction because we'll prove now that
$$xyz+xy+4yz+9zx\leq144.$$ 
Indeed, we need to prove that: $$xyz+\frac{(x+y+z)(xy+4yz+9zx)}{8}\leq\frac{144(x+y+z)^3}{512}$$ or
$$9(x+z)(x-z)^2+y(23x+11z)(x-z)+y^2(9y+23x+11z)\geq0,$$ for which it's enough to prove that
$$y^2(23x+11z)^2-36y^2(x+z)(23x+11z)\leq0,$$ which is obvious.
Id est, $$a+b+c\geq8.$$
The equality occurs for $a=c=4$ and $b=0,$ which says that we got a minimal value.  
