In terms of real parameter $a$ solve inequality:
$\log _{a} {x} + \mid a+\log _{a} {x} \mid \cdot \log _{\sqrt{x}} {a} \geq a\log _{x} {a}$ 
in set of real numbers
1. **Substitute \( \log_a x = t \):**
   \[
   \log_a x = t \implies x = a^t
   \]
   The given inequality becomes:
   \[
   t + |\log_a x + a| \cdot \log_{\sqrt{x}} a \geq a \log_x a
   \]
   Since \( \log_a x = t \), we have:
   \[
   t + |t + a| \cdot \log_{\sqrt{a^t}} a \geq a \log_{a^t} a
   \]

2. **Simplify the logarithmic terms:**
   \[
   \log_{\sqrt{a^t}} a = \log_{a^{t/2}} a = \frac{1}{t/2} = \frac{2}{t}
   \]
   \[
   \log_{a^t} a = \frac{1}{t}
   \]
   Substituting these into the inequality, we get:
   \[
   t + |t + a| \cdot \frac{2}{t} \geq \frac{a}{t}
   \]

3. **Simplify the inequality:**
   \[
   t + \frac{2|t + a|}{t} \geq \frac{a}{t}
   \]
   Multiply through by \( t \) (assuming \( t \neq 0 \)):
   \[
   t^2 + 2|t + a| \geq a
   \]

4. **Consider the cases for \( |t + a| \):**
   - **Case 1: \( t + a \geq 0 \) (i.e., \( t \geq -a \)):**
     \[
     |t + a| = t + a
     \]
     The inequality becomes:
     \[
     t^2 + 2(t + a) \geq a
     \]
     Simplify:
     \[
     t^2 + 2t + 2a \geq a
     \]
     \[
     t^2 + 2t + a \geq 0
     \]
     This is a quadratic inequality. The discriminant of the quadratic equation \( t^2 + 2t + a = 0 \) is:
     \[
     \Delta = 2^2 - 4 \cdot 1 \cdot a = 4 - 4a = 4(1 - a)
     \]
     For the quadratic to be non-negative, the discriminant must be non-negative:
     \[
     4(1 - a) \geq 0 \implies a \leq 1
     \]
     Therefore, for \( a \leq 1 \), the inequality holds for all \( t \geq -a \).

   - **Case 2: \( t + a < 0 \) (i.e., \( t < -a \)):**
     \[
     |t + a| = -(t + a) = -t - a
     \]
     The inequality becomes:
     \[
     t^2 + 2(-t - a) \geq a
     \]
     Simplify:
     \[
     t^2 - 2t - 2a \geq a
     \]
     \[
     t^2 - 2t - 3a \geq 0
     \]
     This is another quadratic inequality. The discriminant of the quadratic equation \( t^2 - 2t - 3a = 0 \) is:
     \[
     \Delta = (-2)^2 - 4 \cdot 1 \cdot (-3a) = 4 + 12a = 4(1 + 3a)
     \]
     For the quadratic to be non-negative, the discriminant must be non-negative:
     \[
     4(1 + 3a) \geq 0 \implies a \geq -\frac{1}{3}
     \]
     Therefore, for \( a \geq -\frac{1}{3} \), the inequality holds for all \( t < -a \).

5. **Combine the results:**
   The inequality holds for:
   \[
   a \leq 1 \quad \text{and} \quad a \geq -\frac{1}{3}
   \]
   Therefore, the solution set for \( a \) is:
   \[
   -\frac{1}{3} \leq a \leq 1
   \]

The final answer is \( \boxed{ -\frac{1}{3} \leq a \leq 1 } \)