Solution:

a. We rewrite the inequality as
$$(\sqrt{x y + 1} + \sqrt{y z + 1} + \sqrt{z x + 1})^2 \leq 2 \cdot (x + y + z)^2$$
and note that, from CBS,
$$\text{LHS} \leq \left(\frac{x y + 1}{x} + \frac{y z + 1}{y} + \frac{z x + 1}{z}\right)(x + y + z)$$
But
$$\frac{x y + 1}{x} + \frac{y z + 1}{y} + \frac{z x + 1}{z} = x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2(x + y + z)$$
which proves (1).

b. The equality occurs when we have equality in CBS, i.e. when
$$\frac{x y + 1}{x^2} = \frac{y z + 1}{y^2} = \frac{z x + 1}{z^2} \left(= \frac{x y + y z + z x + 3}{x^2 + y^2 + z^2}\right)$$
Since we can also write
$$(\sqrt{x y + 1} + \sqrt{y z + 1} + \sqrt{z x + 1})^2 \leq \left(\frac{x y + 1}{y} + \frac{y z + 1}{z} + \frac{z x + 1}{x}\right)(y + z + x) = 2(x + y + z)^2$$
the equality implies also
$$\frac{x y + 1}{y^2} = \frac{y z + 1}{z^2} = \frac{z x + 1}{x^2} \left(= \frac{x y + y z + z x + 3}{x^2 + y^2 + z^2}\right)$$
But then $x = y = z$, and since $x + y + z = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$, we conclude that $x = \frac{1}{x} = 1 = y = z$.

Alternative solution to b): The equality condition
$$\frac{x y + 1}{x^2} = \frac{y z + 1}{y^2} = \frac{z x + 1}{z^2}$$
can be rewritten as
$$\frac{y + \frac{1}{x}}{x} = \frac{z + \frac{1}{y}}{y} = \frac{x + \frac{1}{z}}{z} = \frac{x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}}{x + y + z} = 2$$
and thus we obtain the system:
$$\left\{\begin{array}{l} y = 2x - \frac{1}{x} \\ z = 2y - \frac{1}{y} \\ x = 2z - \frac{1}{z} \end{array}\right.$$
We show that $x = y = z$. Indeed, if for example $x > y$, then $2x - \frac{1}{x} > 2y - \frac{1}{y}$, that is, $y > z$ and $z = 2y - \frac{1}{y} > 2z - \frac{1}{z} = x$, and we obtain the contradiction $x > y > z > x$. Similarly, if $x < y$, we obtain $x < y < z < x$.
Hence, the numbers are equal, and as above we get $x = y = z = 1$.