1. Compare the numbers

$$
a=\left[\log _{2}(\sqrt{5}+1)\right]^{3} \text { and } b=1+\log _{2}(\sqrt{5}+2)
$$

Marius Damian, teacher, Brăila
Solution. For any $x \geq 0$, the inequality $x^{3} \geq 3 x-2$ holds.

Indeed, using the inequality between the arithmetic mean and the geometric mean, we have $x^{3}+2=x^{3}+1+1 \geq 3 \sqrt[3]{x^{3} \cdot 1 \cdot 1}=3 x \Rightarrow x^{3} \geq 3 x-2$.

Equality holds $\Leftrightarrow x=1$.

Using the demonstrated inequality and the fact that $\log _{2}(\sqrt{5}+1)>1$, we have

$$
\begin{gathered}
a=\left[\log _{2}(\sqrt{5}+1)\right]^{3}>3 \log _{2}(\sqrt{5}+1)-2=\log _{2}(\sqrt{5}+1)^{3}-2= \\
=\log _{2}(8 \sqrt{5}+16)-\log _{2} 4=\log _{2} \frac{8 \sqrt{5}+16}{4}=\log _{2}[2(\sqrt{5}+2)]=1+\log _{2}(\sqrt{5}+2)=b
\end{gathered}
$$

In conclusion, $a>b$.