Golden inequality

Since the expression $\frac{(a+b)(b+c)(a+b+c)}{abc}$ is homogeneous in $a,b,c$ , we can assume that $a+b+c = 1$ . Thus we want to minimize $\frac{(1-a)(1-c)}{ac(1-a-c)} \; = \; \frac{1}{ac} + \frac{1}{1-a-c} \; = \; F(a,c)$ over $a,c > 0$ , $a+c < 1$ . Since $\tfrac12(a+c) \ge \sqrt{ac}$ . we deduce that $F(a,c) \ge F\big(\tfrac12(a+c),\tfrac12(a+c)\big)$ and hence we will achieve the minimum value of $F(a,c)$ when $a=c$ . Thus we want to minimize $G(a) \; = \; F(a,a) \; = \; \frac{1}{a^2} + \frac{1}{1-2a} \hspace{2cm} 0 < a < \tfrac12$ Since $G'(a) = \frac{2}{(1-2a)^2} - \frac{2}{a^3}$ , we see that the minimum value of $G$ occurs when $0 = a^3 - (2a-1)^2 = a^3 - 4a^2 + 4a - 1 = (a-1)(a^2 - 3a + 1)$ and hence when $a = \tfrac12(3-\sqrt{5})$ , at which point $G(a) = \boxed{\tfrac12(11+5\sqrt{5})}$ .

I also used a calculus approach. Let $X = \dfrac {(a+b)(b+c)(a+b+c)}{abc}$ . Then consider the value of $a$ in respect to $b$ and $c$ , when $X$ is minimum.

$\frac {\partial X}{\partial a} = \frac {a((b+c)(a+b+c)+(a+b)(b+c) - (a+b)(b+c)(a+b+c)}{a^2bc}$

Putting $\dfrac {\partial X}{\partial a} = 0$ , we have $a^2 = b^2 + bc$ . Similarly, $\dfrac {\partial X}{\partial c} = 0$ , $\implies c^2 = b^2 + ab$ . Since $a$ and $c$ are interchangeable, we can assume $a=c$ when $X$ is minimum. Then

$\begin{aligned} a^2 - ab - b^2 & = 0 \\ \implies a & = \frac {b+\sqrt{b^2 + 4b^2}}2 & \small \blue{\text{for }a > 0} \\ & = \frac {1+\sqrt 5}2 b = \varphi b & \small \blue{\text{where }\varphi \text{ denotes the golden ratio.}} \end{aligned}$

Then minimum value of $X$ :

$\begin{aligned} \min (X) & = \frac {(a+b)(b+c)(a+b+c)}{abc} \\ & = \frac {(\varphi+1)^2 b^2 (2\varphi +1)b}{\varphi^2b^3} \\ & = \frac {\varphi^4 \varphi^3}{\varphi^2} \\ & = \varphi^5 = 5\varphi + 3 \approx \boxed{11.090} \end{aligned}$

Here's a terrible solution. Because I have to resort to calculus, which I don't think is the intended approach.

Without the loss of generality, let $a + b + c = 1$ , then the expression in question can be reduced to a 2-variable expression: $f(a,c) := \dfrac{(1-c)(1-a)}{ac(1-a-c)}$ where $a$ and $c$ are both in the interval $(0,1)$ . Near the boundary points, $f(a,c)$ diverges to infinity, so $f(a,c)$ has no maximum point.

At the minimum point, the partial derivatives of $f(a,c)$ must be 0, $\begin{cases} \dfrac{\delta }{\delta a} f(a,c) = 0 \\ \phantom0 \\ \dfrac{\delta }{\delta c} f(a,c) = 0 \end{cases} \quad\Rightarrow \quad \begin{cases} \dfrac{(c-1)((a-1)^2 - c)}{a^2c (-a-c+1)^2} = 0 \\ \phantom 0 \\ \dfrac{(a-1)((a-1)^2 - c)}{ac^2 (-a-c+1)^2} = 0 \end{cases} \quad \implies \quad \begin{cases} (a-1)^2 - c = 0 \\ (c-1)^2 - a = 0 \end{cases}$ Solving the system of equations above gives $a = c = \dfrac{3-\sqrt5}2 \quad\implies \quad \min[ f(a,c) ] = \boxed{\dfrac{11+5\sqrt5}2} \approx 11.0901699$ Hence, expression in question has a minimum value of $\frac{11+5\sqrt5}2 ,$ when $(a,b,c)$ take the ratio $a:b:c = 2 : (\sqrt5 - 1) : 2 .$

I wonder if there's a way to prove that the minimum value of $f(a,c)$ is $\frac{11+5\sqrt5}2$ without resorting to calculus. That is,

Prove that $2(1-c)(1-a) -(11 + 5\sqrt5) \cdot ac \cdot (1-a-c)$ is strictly non-negative. And it equals 0 when $a=c= \frac{3-\sqrt5}2$ .

I tried the substitution $(a,c) \leftrightarrow (1-a,1-c)$ but I'm still stuck.

Admittedly this is not a very good solution but it spared me a lot of tedious algebra and calculus. I made extensive use of the graphing calculator Desmos to solve this problem.

Since we have three variables that can change, we can try to find when the value of $\frac{(a + b)(b + c)(a + b + c)}{abc}$ is minimized if we only allow one variable at a time to vary. Let $c = x$ and let $f(x) = \frac{(a + b)(b + x)(a + b + x)}{abx}$ . We can expand out $f(x)$ as follows:

$\displaystyle f(x) = \frac{\left(a+b\right)}{ab}\frac{\left(b+x\right)\left(a+b+x\right)}{x}$ $\displaystyle f(x) = \frac{\left(a+b\right)}{ab}\frac{ab+b^{2}+bx+ax+bx+x^{2}}{x}$ $\displaystyle f(x) = \frac{\left(a+b\right)^{2}}{a}\frac{1}{x}+\frac{\left(a+b\right)}{ab}\left(a+2b+x\right)$

Taking the derivative of $f(x)$ and setting find the $x$ such that $f'(x) = 0$ gives the equation $\displaystyle -\frac{\left(a+b\right)^{2}}{a}\frac{1}{x^{2}}+\frac{\left(a+b\right)}{ab}=0$ or $x=\sqrt{\left(a+b\right)b}$ .

Here is where desmos comes in. If we let $c = \sqrt{\left(a+b\right)b}$ , we can graph $g(x) = \frac{(a + x)(x + c)(a + x + c)}{axc}$ against $x$ and vary the value of $a$ in the interval $(0, \infty)$ . It just so happens that the minimum of this function always occurs at the same y-value independent of our choice for $a$ (so long as $a$ is positive and real). The minimum of $g(x)$ is approximately $\boxed{11.09017}$ .