For what value of the parameter $p$ does the following inequality hold for all positive $x$:

$$
\lg (x+p)-\frac{1}{2} \geq \lg \sqrt{2 x}
$$
Using the properties of the logarithmic function and the fact that it is strictly monotonically increasing for bases greater than 1, the given inequality is equivalent to:

$$
\frac{x+p}{\sqrt{10}} \geq \sqrt{2 x}
$$

By squaring and rearranging, we arrive at the inequality:

$$
x^{2}+(2 p-20) x+p^{2} \geq 0
$$

The question is for which values of $p$ this inequality holds for all positive $x$. Since the value of $x^{2}+(2 p-20) x+p^{2}$ at $x=0$ is $p^{2} \geq 0$, the inequality holds if the parabola of $x^{2}+(2 p-20) x+p^{2}$

a) has its vertex in the half-plane $x \leq 0$;

b) has its vertex above the $x$-axis.

The vertex is located at $-\frac{2 p-20}{2}$, so the condition a) is equivalent to $p \geq 10$. The vertex is above the $x$-axis if the discriminant is negative (since the coefficient of $x^{2}$ is positive), i.e., $(2 p-20)^{2}-4 p^{2} \leq 0$, so condition b) is equivalent to $p \geq 5$.

Thus, the inequality holds for all positive $x$ if $p \geq 5$.