Let

$$
A=\sqrt{5}+\sqrt{22+2 \sqrt{5}}, \quad B=\sqrt{11+2 \sqrt{29}}+\sqrt{16-2 \sqrt{29}+2 \sqrt{55-10 \sqrt{29}}}
$$

What is the magnitude relationship between the numbers $A$ and $B$?
To speak about the magnitude of the given quantities, we must show that these quantities exist, that is, there are no negative numbers under the square root signs. This is not obvious for the second term of the sum forming $B$. Here, $\sqrt{55-10 \sqrt{29}}$ is defined, since $55>10 \sqrt{29}$ (dividing by 5 and squaring gives $121>116$). By transforming the quantity under the "large" square root sign, we get

$$
16-2 \sqrt{29}+2 \sqrt{55-10 \sqrt{29}}=5+(11-2 \sqrt{29})+2 \sqrt{5} \sqrt{11-2 \sqrt{29}}
$$

and from here it is visible that we are dealing with the square of the sum of two terms, $\sqrt{5}$ and $\sqrt{11-2 \sqrt{29}}$. Accordingly,

$$
B=\sqrt{11+2 \sqrt{29}}+\sqrt{(\sqrt{5}+\sqrt{11-2 \sqrt{29}})^{2}}=\sqrt{5}+\sqrt{11+2 \sqrt{29}}+\sqrt{11-2 \sqrt{29}}
$$

The magnitude of the two quantities now depends on whether

$$
A_{1}=\sqrt{22+2 \sqrt{5}} \text { and } B_{1}=\sqrt{11+2 \sqrt{29}}+\sqrt{11-2 \sqrt{29}}
$$

which is the larger. Squaring gives us that

$$
A_{1}^{2}=22+2 \sqrt{5}, \quad B_{1}^{2}=22+2 \sqrt{11^{2}-(2 \sqrt{29})^{2}}=22+2 \sqrt{5}
$$

We have found that $A_{1}^{2}=B_{1}^{2}$ and thus $A_{1}=B_{1}$, since we are dealing with positive numbers, so the numbers $A$ and $B$ are equal.