Minimum value when $abc+ab+4bc+9ca=144$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

[Hanno's CS solution is much more instructive, so read that. I was pleasantly surprised that this approach worked, so I posted it.]

1/ First, suppose $c$ is fixed, and we want to minimize $a+b$. The condition is equivalent to

$$ (c+1) ab + (4c) b + (9c) a = 144, $$

which we factorize (via Simon's Favorite Factoring Trick) to

$$[ (c+1 ) a + 4c ] [ (c+1) b + 9c ] = 144(c+1) + 36 c^2, $$

and applying AM-GM on the LHS (both terms are non-negative), we get that

$$ [ (c+1 ) a + 4c ] + [ (c+1) b + 9c ] \geq 2 \sqrt{ 144(c+1) + 36 c^2 } = 12(c+2), $$

which simplifies to (We divide by $ c+1 \geq 0$)

$$ a+b \geq \frac{ -c + 24 } { c+1} , $$

or that

$$ a + b +c \geq \frac{ c^2 + 24 } { c+1}. $$

Equality holds when $ (c+1) a + 4c = (c+1)b + 9c $, or that $ a-b = \frac{5c}{c+1}$.

2/ Show that on $ c \geq 0 $, we have $ \frac{ c^2 + 24 } { c+1 } \geq 8 $ with equality at $ c = 4$.
This can be done through differentiation or through clever AM-GM (if you want to avoid calculus).

3/ Hence, conclude that $ a+b+c \geq 8$, with equality when $c = 4, a - b = 4$. We substitute this into the original condition, and solve the quadratic to get the equality case of $ a = 4, b = 0, c = 4$ (reject negative cases ).

Notes

• There was nothing special about fixing $c$. We could also have proceeded with $a$ or $b$ fixed, and arrived at the same conclusion.
  • In fact, if we did the similar calculations, we could determine the values of $a = 4, b = 0, c = 4$, and then verify that the original condition is satisfied.
• Hanno's observation about transforming to remove coefficients would have simplified this writeup.