Assume that real numbers $a$ and $b$ satisfy

$$
a b+\sqrt{a b+1}+\sqrt{a^{2}+b} \cdot \sqrt{b^{2}+a}=0
$$

Find, with proof, the value of

$$
a \sqrt{b^{2}+a}+b \sqrt{a^{2}+b}
$$
Let us rewrite the given equation as follows:

$$
a b+\sqrt{a^{2}+b} \sqrt{b^{2}+a}=-\sqrt{a b+1} .
$$

Squaring this gives us

$$
\begin{aligned}
a^{2} b^{2}+2 a b \sqrt{a^{2}+b} \sqrt{b^{2}+a}+\left(a^{2}+b\right)\left(b^{2}+a\right) & =a b+1 \\
\left(a^{2} b^{2}+a^{3}\right)+2 a b \sqrt{a^{2}+b} \sqrt{b^{2}+a}+\left(a^{2} b^{2}+b^{3}\right) & =1 \\
\left(a \sqrt{b^{2}+a}+b \sqrt{a^{2}+b}\right)^{2} & =1 \\
a \sqrt{b^{2}+a}+b \sqrt{a^{2}+b} & = \pm 1 .
\end{aligned}
$$

Next, we show that \( a \sqrt{b^{2}+a}+b \sqrt{a^{2}+b} > 0 \). Note that

$$
a b=-\sqrt{a b+1}-\sqrt{a^{2}+b} \cdot \sqrt{b^{2}+a} < 0 \implies a b < 0 \implies a < 0 \text{ and } b > 0 \text{ or } a > 0 \text{ and } b < 0.
$$

Without loss of generality, assume \( a < 0 \) and \( b > 0 \). Then rewrite

$$
a \sqrt{b^{2}+a}+b \sqrt{a^{2}+b}=a\left(\sqrt{b^{2}+a}+b\right)-b\left(a-\sqrt{a^{2}+b}\right)
$$

and, since \( \sqrt{b^{2}+a}+b \) and \( a-\sqrt{a^{2}+b} \) are both positive, the expression above is positive. Therefore,

$$
a \sqrt{b^{2}+a}+b \sqrt{a^{2}+b}=1,
$$

and the proof is finished.