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?
 A: 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.  
A: You can get there without resorting on Lagrange multipliers.
First the constraint $\,144=abc+ab+4bc+9ca\,$ is transformed into a more suitable expression:
Divide it by $36$ and rescale variables by $\,\alpha=\frac a4, \beta=\frac b9,$ and $\gamma=c.$ This leads to
$$\begin{align*}4 \:=\:\alpha\beta\gamma & +\alpha\beta+\beta\gamma + \gamma\alpha\quad
\big| +\alpha\beta +\beta\gamma+\gamma\alpha +4(\alpha+\beta+\gamma) +8 \\[1.6ex]
\iff\;\sum_{\text{cyc}}(\alpha+2)(\beta+2) & \:=\:(\alpha+2)(\beta+2)(\gamma+2)\quad\Big|\:\cdot\frac1{\text{RHS}}\\
\iff\;\sum_{\text{cyc}}\frac1{\alpha+2} & \:=\:1\tag{1}
\end{align*}$$
Next apply the Cauchy–Bunyakovsky–Schwarz inequality and exploit $(1)$ to obtain
$$\begin{align*}
(2+3+1)^2 & \:=\:
\left(2\sqrt{\alpha+2}\cdot\frac1{\sqrt{\alpha+2}}
\,+\,3\sqrt{\beta+2}\cdot\frac1{\sqrt{\beta+2}}
\,+\,\sqrt{\gamma+2}\cdot\frac1{\sqrt{\gamma+2}}\right)^2 \\[1ex]
& \:\leqslant\: 4(\alpha+2)+9(\beta+2)+\gamma+2 \\[2ex]
\iff\quad 8 & \:\leqslant\:4\alpha+9\beta+\gamma \:=\:a+b+c
\end{align*}$$
Finally, one has equality only if one argument vector is a scalar multiple of the other. Thus,
$$2\sqrt{\alpha+2}=\frac\lambda{\sqrt{\alpha+2}}\;
\text{ and so on, or }\;2(\alpha+2)=\lambda=3(\beta+2) = \gamma+2\,,$$ 
which yields $\lambda=6$ using $(1)$. Hence $(a,b,c)=(4,0,4)\,$
is the unique minimising solution.
