Awesome Inequality

Let $P=\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\right)$

We have:

$\begin{aligned} 2P&=\left(\sum\dfrac{x}{y}+\sum\dfrac{y}{x}\right)^2-\left(\sum\dfrac{x}{y}\right)^2-\left(\sum\dfrac{y}{x}\right)^2\\ &=64-\left(\sum\dfrac{x^2}{y^2}+2\sum\dfrac{x}{y}\right)+\left(\sum\dfrac{y^2}{x^2}+2\sum\dfrac{y}{x}\right)\\ &=48-\left(\sum\dfrac{x^2}{y^2}+\sum\dfrac{y^2}{x^2}\right)\\ &=51-\left(\sum x^2\right)\left(\sum\dfrac{1}{x^2}\right) \end{aligned}$

From the hypothesis, we get: $\displaystyle\left(\sum x\right)\left(\sum\dfrac{1}{x}\right)=\left(\sum xy\right)\left(\sum\dfrac{1}{xy}\right)=11$

Applying the Cauchy-Schwarz inequality we have:

$\begin{aligned} &\sqrt{\left(2\sum xy\right)\left(2\sum\dfrac{1}{xy}\right)}+\sqrt{\left(\sum x^2\right)\left(\sum\dfrac{1}{x^2}\right)}\\ \le&\sqrt{\left(2\sum xy+\sum x^2\right)\left(2\sum\dfrac{1}{xy}+\sum\dfrac{1}{x^2}\right)}\\ =&\left(\sum x\right)\left(\sum\dfrac{1}{x}\right) \end{aligned}$

Hence, $\displaystyle \left(\sum x^2\right)\left(\sum\dfrac{1}{x^2}\right)\le \left(11-2\sqrt{11}\right)^2$

So, $\displaystyle P\ge\dfrac{51- \left(11-2\sqrt{11}\right)^2}{2}=22\sqrt{11}-57$

Thus, $a+b+c=\boxed{-24}$ .

Write $a = x + y + z$ , $b = xy + xz = yz$ and $c = xyz$ . Then $8 \; = \; \left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x}\right) + \left(\frac{x}{z} + \frac{z}{y} + \frac{y}{x}\right) \; = \; \frac{x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2}{xyz} \; = \; \frac{ab - 3c}{c}$ and hence $ab \; = \; 11c \;$ and so $x,y,z$ are the roots of the cubic $f(X) \; = \; X^3 - aX^2 + bX - \tfrac{1}{11}ab \; = \; 0 \;.$ Playing with elementary symmetric polynomials gives $\begin{array}{rcl} \mathcal{Q} & = & \left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x}\right)\left(\frac{x}{z} + \frac{z}{y} + \frac{y}{x}\right) \\ & = & 3 + \left(\frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy}\right) + \left(\frac{yz}{x^2} + \frac{xz}{y^2} + \frac{xy}{z^2}\right) \\ & = & 3 + \frac{x^3 + y^3 + z^3}{c} + \frac{x^3y^3 + x^3z^3 + y^3z^3}{c^2} \\ & = & 3 + \frac{a^3 - 3ab + 3c}{c} + \frac{b^3 - 3abc + 3c^2}{c^2} \\ & = & 9 + \frac{a^3}{c} + \frac{b^3}{c^2} - \frac{6ab}{c} \\ & = & 9 + \frac{11a^2}{b} + \frac{121b}{a^2} - 66 \\ & = & 11u + 121u^{-1} - 57 \end{array}$ where $u \,=\, a^2b^{-1}$ .

Since $\mathcal{Q}'(u) \,=\, 11 - 121u^{-2}$ , we see that the minimum of $\mathcal{Q}$ over all positive $u$ occurs when $u = \sqrt{11}$ , giving a value of $\mathcal{Q} = 22\sqrt{11} - 57$ .

It remains to show that these values of $u$ and $\mathcal{Q}$ can be achieved with $x,y,z$ all positive. With $a = b = \sqrt{11}$ , we have the cubic $f(X) \; = \; X^3 - \sqrt{11} X^2 + \sqrt{11} X - 1 \; = \; (X - 1)\big(X^2 + (1 - \sqrt{11})X + 1\big) \;,$ which has positive roots $1$ , $\tfrac12\big[\sqrt{11}-1 \pm \sqrt{8 - 2\sqrt{11}}\big]$ . Thus this value of $\mathcal{Q}$ can be achieved, and so the desired answer is $22 + 11 - 57 =\boxed{-24}$ .