How to solve the inequality analytically? 
Solve the inequality analytically.
  $$ \log_x{(\sqrt{x^2 + 2x - 3} + 2)} \cdot \log_5{(x^2 + 2x - 2)} \ge \log_x{4} $$

My solution
$$
\left.
\left\{
    \begin{array}{l}
        x > 0 \\
        x \ne 1 \\
        x^2 + 2x - 3 \ge 0 \\
        x^2 + 2x - 2 > 0
    \end{array}
\right.
\right\vert \Rightarrow x \in (1; + \infty).
$$
Since $\log_a{b} = \frac{\log_c{b}}{\log_c{a}}$,
$$
\frac{\log_5(\sqrt{x^2 + 2x - 3} + 2) \cdot \log_5{(x^2 + 2x -2)}}{\log_5{x}}
\ge
\frac{\log_5{4}}{\log_5{x}}
$$
$$
\log_5{(\sqrt{x^2 + 2x - 3} + 2)} \cdot \log_5{(x^2 + 2x - 2)} \ge \log_5{4}.
$$
Let $b = x^2 + 2x$,
$$
\log_5{(\sqrt{b - 3} + 2)} \cdot \log_5{(b - 2)} \ge \log_5{4}.
$$
Unfortunately, that's all that came to mind.
P.S. Excuse me for my bad English! It isn't my native language.

You have correctly eliminated $x \leqslant 1$.  For $x> 1$, we may take powers of $x$ on both sides and have the equivalent:
$$(x^2+2x-3)^{\log_5(x^2+2x-2)} \geqslant 4$$
Now if $b=x^2+2x-2 \in (1, 5)$, then in the LHS, the exponent is $\in (0, 1)$, while the base is $\in (0, 4)$. Clearly the LHS will be less than RHS here.
We are left with $b \geqslant 5$ which gives exponent $ \geqslant 1$ and base $\geqslant 4$, which clearly, will always work. Thus what is left to solve is $x^2+2x-2\geqslant 5$, which is left for you...
