Solve the equation $\frac{7x^2 - x + 4}{\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1}} = 2\sqrt 2$. 
Solve the equation $\dfrac{7x^2 - x + 4}{\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1}} = 2\sqrt 2$ over the reals.

Remember when I posted a question almost identical to this one?
Well, turned out my teacher actually mistyped the question. But even when the problem is correctly typed, we couldn't still figure it out throughout the lesson. (And remember the solution must be able to be executed in a test setting.)
 A: Using the Bunyakovsky and AM - GM inequalities for $(\sqrt{3x^2 - 1}, \sqrt{x^2 - x}, \sqrt{x^2 + 1})$, $(1, 1, -x)$ and $(2x^2 + 4)$, $(5x^2 - x)$, we have that $$(\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1})^2 \le (1 + 1 + x^2)(3x^2 - 1 + x^2 - x + x^2 + 1)$$
$$ = (x^2 + 2)(5x^2 - x) = \dfrac{(2x^2 + 4)(5x^2 - x)}{2} \le \dfrac{(7x^2 - x + 4)^2}{8}$$
The equality sign occurs when $\sqrt{3x^2 - 1} = \sqrt{x^2 - x} = -\dfrac{\sqrt{x^2 + 1}}{x}$ and $2x^2 + 4 = 5x^2 - x$.
From the second condition, we have that $3x^2 - x - 4 = 0 \implies \left [ \begin{align} x &= -1\\ x &= \dfrac{4}{3} \end{align} \right.$.
Plugging in $x = -1$, we have that $\sqrt{3x^2 - 1} - \sqrt{x^2 - x} = \sqrt{3 \cdot 1^2 - 1} - \sqrt{1^2 + 1} = 0$.
And $-\dfrac{2\sqrt{x^2 + 1}}{x} - \sqrt{3x^2 - 1} = \dfrac{2\sqrt{1^2 + 1}}{1} - \sqrt{3 \cdot 1^2 - 1} = 0$
Plugging in $x = \dfrac{4}{3}$, we have that $\sqrt{3x^2 - 1} - \sqrt{x^2 - x} = \sqrt{3 \cdot \left(\dfrac{4}{3}\right)^2 - 1} - \sqrt{\left(\dfrac{4}{3}\right)^2 - \dfrac{4}{3}} \ne 0$.
So $x = -1$.
