## Task B-3.1.

Solve the inequality

$$
\sqrt{4^{x}+1} \geqslant\left|4^{x-1}-1\right|+4^{x} \log _{x} \sqrt{x}
$$

in the set of real numbers.
## Solution.

For all expressions to be defined, it must hold that $x>0$ and $x \neq 0$. Under this condition and the properties of logarithms, the given inequality is equivalent to the inequalities

$$
\begin{aligned}
& \sqrt{4^{x}+1} \geqslant\left|4^{x-1}-1\right|+4^{x} \cdot \frac{1}{2} \\
& \sqrt{4^{x}+1} \geqslant\left|\frac{1}{4} \cdot 4^{x}-1\right|+\frac{1}{2} \cdot 4^{x}
\end{aligned}
$$

Introduce the substitution $t=4^{x}$. Since $x>0, x \neq 1$, it follows that $t>1$ and $t \neq 4$.

This leads to $\sqrt{t+1} \geqslant\left|\frac{1}{4} t-1\right|+\frac{1}{2} t$,

and after multiplying by 4, we get

$$
\sqrt{16 t+16} \geqslant|t-4|+2 t
$$

First case. Let $10$, squaring is allowed, so we have $16 t+16 \geqslant t^{2}+8 t+16$, which simplifies to $t^{2}-8 t \leqslant 0$.

Since $t>1$, we get $t \leqslant 8$, meaning that every $t$ for which $14$ holds.

If $t>4$, then $3 t-4>0$ and squaring is allowed.

We get $16 t+16 \geqslant 9 t^{2}-24 t+16$, which simplifies to $9 t^{2}-40 t \leqslant 0$. This leads to $9 t-40 \leqslant 0$ and thus $4<t \leqslant \frac{40}{9}$.

Then $4<4^{x} \leqslant \frac{40}{9}$, or $1<x \leqslant \log _{4} \frac{40}{9}$.

The final solution is $x \in\langle 0,1\rangle \cup\left\langle 1, \log _{4} \frac{40}{9}\right]$.