7. Given $a, b>0$. Then the minimum value of $\frac{b^{2}+2}{a+b}+\frac{a^{2}}{a b+1}$ is . $\qquad$
7. 2 .

By Cauchy-Schwarz inequality,
$$
\begin{array}{l}
\frac{b^{2}+2}{a+b}+\frac{a^{2}}{a b+1} \\
\geqslant \frac{b^{2}+2}{\sqrt{a^{2}+1} \sqrt{b^{2}+1}}+\frac{a^{2}}{\sqrt{a^{2}+1} \sqrt{b^{2}+1}} \\
=\frac{\left(a^{2}+1\right)+\left(b^{2}+1\right)}{\sqrt{a^{2}+1} \sqrt{b^{2}+1}} \geqslant 2,
\end{array}
$$

The equality holds when $(a, b)=(1,1)$.