What is the largest real number M satisfying a specific inequality?

[anemone]

Find the largest possible real number $M$ such that for all pairs $(a,\,b)$ of real numbers with $a\ne b$, and $ab=2$,

$\dfrac{((a+b)^2-6)((a-b)^2+8)}{(a-b)^2}\ge M$.

Also, determine for which pairs $(a,\,b)$ equality holds.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. Opalg
2. lfdahl
3. greg1313

Solution from Opalg:

Spoiler

If $ab=2$ then $(a\pm b)^2 = a^2 \pm 2ab + b^2 = a^2+b^2 \pm4 = x \pm4$, where $x = a^2+b^2$. Therefore $$\frac{\bigl((a+b)^2 -6\bigr) \bigl((a-b)^2 + 8\bigr)}{(a-b)^2} = \frac{(x-2)(x+4)}{x-4} = x-4 + \frac{16}{x-4} + 10$$ (notice that $x>4$, because $2 = ab < \frac12(a^2+b^2) = \frac12x$).

The minimum value of $x-4 + \frac{16}{x-4}$ occurs when $x-4 = \sqrt{16} = 4$, so that $a^2 + b^2 = 8$, giving $M = 4 + \frac{16}4 + 10 = 18.$

For equality to hold, we must have $a^2 + b^2 = 8$ and $ab=2$, so that $(a+b)^2 = a^2+b^2 + 2ab = 12$ and $a+b = \pm2\sqrt3.$

Thus $a$ and $b$ are the roots of the equation $\lambda^2 \pm 2\sqrt3\lambda + 2$, namely $\pm\sqrt3 \pm1$. So the pairs $(a,b)$ for which equality holds are $(\sqrt3+1,\sqrt3-1)$, $(\sqrt3-1,\sqrt3+1)$, $(-\sqrt3+1,-\sqrt3-1)$, $(-\sqrt3-1,-\sqrt3+1)$.