Problem 1504

[Proof] Since $a \in(0,1), b \in(0,1)$, we can set
$$a=\cos \alpha, b=\cos \beta$$

where $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$. Clearly, using the above expressions, we get
$$\begin{aligned} a b+\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)} & =\cos \alpha \cos \beta+\sin \alpha \sin \beta \\ & =\cos (\alpha-\beta) . \end{aligned}$$

Squaring both sides of the above equation, we have
$$\begin{aligned} & a^{2} b^{2}+2 a b \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}+\left(1-a^{2}\right)\left(1-b^{2}\right) \\ = & \cos ^{2}(\alpha-\beta) \end{aligned}$$

Rearranging terms, we get
$$2 a b \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}=\cos ^{2}(\alpha-\beta)-1+a^{2}+b^{2}-2 a^{2} b^{2}$$

Dividing both sides by $2 a b$, we have
$$\sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}=\frac{1}{2 a b}\left(\cos ^{2}(\alpha-\beta)-1\right)+\frac{a}{2 b}+\frac{b}{2 a}-a b .$$

Noting that when $0 < |\alpha - \beta| < \frac{\pi}{6}$, we have
$$\cos (\alpha - \beta) > \frac{\sqrt{3}}{2}$$

Thus,
$$\cos ^{2}(\alpha-\beta)-1>\frac{3}{4}-1=-\frac{1}{4}$$

At this point, the inequality in the problem holds.
Therefore, within the open interval $\left(0, \frac{\pi}{2}\right)$, choose 4 pairs of distinct angles $(\alpha, \beta)$ such that $0 < |\alpha - \beta| < \frac{\pi}{6}$, and then take 4 pairs of distinct positive numbers $(a, b)$, where $a=\cos \alpha$, $b=\cos \beta$, so that the inequality given in the problem holds.