2. Find the minimum value of the expression for $a, b>0$

$$
\frac{(2 a+2 a b-b(b+1))^{2}+\left(b-4 a^{2}+2 a(b+1)\right)^{2}}{4 a^{2}+b^{2}}
$$
Answer: 1.

Solution 1. The minimized expression is the sum of the squares of the distances from point $A(1,2 a)$ to the lines $\ell_{1}$ and $\ell_{2}$, given by the equations

$$
\ell_{1}: 2 a x+b y-b(b+1)=0 \quad \text { and } \quad \ell_{2}: b x-2 a y+2 a(b+1)=0
$$

These lines are perpendicular and intersect at point $B(0, b+1)$. Then, by the Pythagorean theorem, we need to find the minimum of $A B^{2}=1+(b+1-2 a)^{2}$, which is 1 and is achieved, for example, when $a=b=1$.

Solution 2. Transform the numerator of the fraction:

$$
\begin{aligned}
& (2 a+2 a b-b(b+1))^{2}+\left(b-4 a^{2}+2 a(b+1)\right)^{2}= \\
& =4 a^{2}+4 a^{2} b^{2}+b^{2}(b+1)^{2}+8 a^{2} b-4 a b(b+1)-4 a b^{2}(b+1)+ \\
& +b^{2}+16 a^{4}+4 a^{2}(b+1)^{2}-8 a^{2} b+4 a b(b+1)-16 a^{3}(b+1)= \\
& \quad=\left(4 a^{2}+b^{2}\right)\left(1+4 a^{2}+(b+1)^{2}-4 a(b+1)\right)=\left(4 a^{2}+b^{2}\right)\left(1+(b+1-2 a)^{2}\right)
\end{aligned}
$$

Thus, we need to find the minimum of the function $(b+1-2 a)^{2}+1$ for positive $a$ and $b$. It is 1 and is achieved, for example, when $a=b=1$.