Intervals

First note that for the terms to be real-valued we must have that

$(2x - 1) \ge 0 \Longrightarrow x \ge \frac{1}{2}$ .

Now, square both sides to find that

$(x + \sqrt{2x - 1}) + (x - \sqrt{2x - 1}) + 2\sqrt{x^{2} - (2x - 1)} = 2$

$\Longrightarrow 2x + 2\sqrt{(x - 1)^{2}} = 2$

$\Longrightarrow 1 - x = \sqrt{(x - 1)^{2}}$

$\Longrightarrow 1 - x = |x - 1|$ .

Now for $x \le 1$ we have that $|x - 1| = -(x - 1) = 1 - x$ , so for $x \le 1$ we end up with the tautology $1 - x = 1 - x$ , i.e., true for any $x \le 1$ .

For $x \ge 1$ we would end up with $1 - x = x - 1 \Longrightarrow x = 1$ , a solution we already know exists.

Combining this with our initial observation gives us that $\frac{1}{2} \le x \le 1$ .

Thus $a = 1, b = 2, c = 1$ and $\frac{b}{a + c} = \frac{2}{1 + 1} = \boxed{1}$ .

a = 1, b = 2 and c = 1