5. Solve the inequality $x+\sqrt{x^{2}+4} \geq\left(2 x+1+\sqrt{4 x^{2}+4 x+2}\right) \cdot 8^{x+1}$.
Answer: $x \in\left(-\infty ;-\frac{2}{3}\right]$.

Solution. The original inequality is equivalent to the inequality:

$$
2\left(\frac{x}{2}+\sqrt{\left(\frac{x}{2}\right)^{2}+1}\right) \geq\left(2 x+1+\sqrt{(2 x+1)^{2}+1}\right) \cdot 8^{x+1} \Leftrightarrow \frac{x}{2}+\sqrt{\left(\frac{x}{2}\right)^{2}+1} \geq\left(2 x+1+\sqrt{(2 x+1)^{2}+1}\right) \cdot 2^{3 x+2}
$$

The function $f(t)=t+\sqrt{t^{2}+1}$ is positive and monotonically increasing everywhere. This is obvious for $t \geq 0$, and for $t<0$, it is also true. If $x>-\frac{2}{3}$, then $\frac{x}{2}>2x+1$, and since $f\left(\frac{x}{2}\right)>f(2 x+1)>0$, the inequality $2^{3 x+2}\left(2 x+1+\sqrt{(2 x+1)^{2}+1}\right) \cdot 2^{3 x+2}$ does not hold.

Thus, $x \leq-\frac{2}{3}$.

Answer to variant 172: $x \in\left[-\frac{3}{8} ;+\infty\right)$.

Answer to variant 173: $x \in\left(-\infty ; \frac{2}{3}\right]$.

Answer to variant 174: $x \in\left[\frac{3}{8} ;+\infty\right)$.