An algebra problem by Anirudha Brahma

$\text {Let}$ ,

$a+2b+c = x$

$a+b+2c = y$

$a+b+3c = z$

$\text {Solving for a,b,c we get}$

$c = z-y$

$b = x+z-2y$

$a = 5y-x-3z$

$\dfrac{5y-x-3z+3z-3y}{x} + \dfrac{4(x+z-2y)}{y} - \dfrac{8(z-y)}{z}$

$\dfrac{2y}{x} -1 + \dfrac{4x}{y} + \dfrac{4z}{y} -8 -8 +\dfrac{8y}{z}$

$-17 + (\dfrac{2y}{x} + \dfrac{4x}{y} + \dfrac{4z}{y} + \dfrac{8y}{z})$

$\text {Applying AM. GM. Inequality to first two fractions}$

$\dfrac{\dfrac{2y}{x} + \dfrac{4x}{y}}{2} \geq \sqrt{8}$

$\dfrac{2y}{x} + \dfrac{4x}{y} \geq 4 \sqrt{2} -(i)$

$\text {Applying AM. GM. Inequality to last two fractions}$

$\dfrac{\dfrac{4z}{y} + \dfrac{8y}{z}}{2} \geq \sqrt{32}$

$\dfrac{2y}{x} + \dfrac{4x}{y} \geq 4 \sqrt{2} - (ii)$

$\text{Minimum value of the expression = -17 + (i) + (ii)}$

$= -17 + 12 \sqrt{2}$

$a+b = -17 + 12$

$\boxed{-5}$

Note that the quantity we are minimizing is homogeneous in $a,b,c$ , so we can assume that $a+b+c=1$ . Since the quantity we are trying to minimize is a continuous function of $a,b,c$ over the compact region $a,b,c,\ge 0$ , $a+b+c=1$ , there must be a minimum value!

We need to minimize $X \; = \; \frac{1-b+2c}{1+b} + \frac{4b}{1+c} - \frac{8c}{1+2c} \; = \; F(b,c) \; = \; \frac{2(1+c)}{1+b} + \frac{4b}{1+c} + \frac{4}{1+2c} - 5$ for $a+b+c = 1$ , so for $b+c \le 1$ . Solving for $\nabla F =\mathbf{0}$ we see that $\frac{\partial F}{\partial b} \; = \; -\frac{2(1+c)}{(1+b)^2} + \frac{4}{1+c} \qquad \qquad \frac{\partial F}{\partial c} \; = \; \frac{2}{1+b} - \frac{4b}{(1+c)^2} - \frac{8}{(1+2c)^2}$ The vanishing of $\frac{\partial F}{\partial b}$ tells us that $(1+c)^2 = 2(1 + b)^2$ , and hence $1+c = \sqrt{2}(1+b)$ . Putting this result into the requirement that $\frac{\partial F}{\partial c} = 0$ tells us that $(1+2c)^2 = 2(1+c)^2$ , and hence $c = 1/\sqrt{2}$ . Thus $b = \tfrac12(\sqrt{2}-1)$ , and hence $a = \tfrac12(3 - 2\sqrt{2})$ . The value of $X$ for these values of $a,b,c$ is $12\sqrt{2}-17$ .

I am not going into the details of determining the Hessian matrix at this point to show that we have a minimum for $F$ ; nor am I going to show that the value of $F$ on the boundary of the region $b,c \ge 0$ , $b+c \le 1$ is greater than $12\sqrt{2}-17$ ; this is left to the interested reader!

The end result is that the answer is correct, but for a much more complicated reason than originally assumed!