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...