Solve the Logarithmic Inequality: log₄x × log₅64 ≥ log₅(x³+x²+x+1)

To solve this problem, we need to compare the expressions $\log_4 x \times \log_5 64$ and $\log_5 (x^3 + x^2 + x + 1)$.

First, calculate $\log_5 64$. We know that $64 = 4^3 = 2^6$. Therefore:
$\log_5 64 = \frac{\log_5 2^6}{\log_5 4} = \frac{6 \log_5 2}{2 \log_5 2} = 3$

Next, simplify the left-hand side expression $\log_4 x$. Using the change of base formula:
$\log_4 x = \frac{\log_5 x}{\log_5 4}$

Therefore, the left-hand side becomes:
$\frac{\log_5 x}{\log_5 4} \times 3 = \frac{3 \log_5 x}{2 \log_5 2}$

For the inequality:
$\frac{3 \log_5 x}{2 \log_5 2} \ge \log_5 (x^3 + x^2 + x + 1)$

We can now equate the right-hand side:
$\log_5 x^{3/2\log_5 2} \ge \log_5 (x^3 + x^2 + x + 1)$

This implies:
$x^{3/2\log_5 2} \ge x^3 + x^2 + x + 1$

Testing and analyzing this expression results in no valid $x$ satisfying the inequality within real values since exponential growth and polynomial terms do not align. Thus, the inequality cannot be satisfied, and no solution satisfies the given conditions.

Therefore, the solution to the problem is: No solution.