Problem 1778

39. Since the degrees at both ends are equal, we might as well set $a b+b c+c a=1$, then let $a=\tan \alpha, b=\tan \beta, c=\tan \gamma$, where $\alpha+\beta+\gamma=90^{\circ}$. $\alpha, \beta, \gamma$ are all acute angles. Then the problem is reduced to proving
$$(b+c)(c+a)(a+b) \geqslant \frac{8 \sqrt{3}}{9}$$

Since $\alpha+\beta+\gamma=90^{\circ}$, we have
$$\begin{array}{l} \sin (\alpha+\beta)=\cos \gamma \\ \sin (\beta+\gamma)=\cos \alpha \\ \sin (\gamma+\alpha)=\cos \beta \\ (b+c)(c+a)(a+b) \geqslant \frac{8 \sqrt{3}}{9} \Leftrightarrow \\ (\tan \alpha+\tan \beta)(\tan \beta+\tan \gamma)(\tan \gamma+\tan \alpha) \geqslant \frac{8 \sqrt{3}}{9} \Leftrightarrow \\ \frac{\sin (\alpha+\beta) \sin (\beta+\gamma) \sin (\gamma+\alpha)}{\cos ^{2} \alpha \cos ^{2} \beta \cos ^{2} \gamma} \geqslant \frac{8 \sqrt{3}}{9} \Leftrightarrow \\ \cos \alpha \cos \beta \cos \gamma \leqslant \frac{3 \sqrt{3}}{8} \end{array}$$

Below are two methods to prove (2).
Proof 1 Since $\cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha-\beta)+\cos (\alpha+\beta)] \leqslant \frac{1}{2}(1+\cos (\alpha+\beta))=\frac{1}{2}(1+\sin \gamma)$, we have,
$$\begin{array}{l} \cos \alpha \cos \beta \cos \gamma \leqslant \frac{1}{2}(1+\sin \gamma) \cos \gamma=\frac{1}{2} \sqrt{(1+\sin \gamma)^{2}\left(1-\sin ^{2} \gamma\right)}= \end{array}$$
$$\begin{array}{l} \frac{1}{2} \sqrt{\frac{1}{3}\left(\frac{3(1+\sin \gamma)+(3-3 \sin \gamma)}{4}\right)^{4}}=\frac{3 \sqrt{3}}{8} \end{array}$$

Equality holds if and only if $\boldsymbol{\alpha}=\boldsymbol{\beta}=\gamma=30^{\circ}$.
Proof 2 Considering that equality in inequality (2) holds if and only if $\alpha=\beta=\gamma=30^{\circ}$, we aim in this direction for adjustment.
By symmetry, we may assume $\alpha \leqslant \beta \leqslant \gamma$, hence $\gamma \geqslant 30^{\circ}, \alpha \leqslant 30^{\circ}$, we first prove
$$\cos \alpha \cos y \leqslant \cos 30^{\circ} \cos \left(\alpha+\gamma-30^{\circ}\right)$$

That is
$$\begin{array}{l} \cos (\alpha-\gamma)+\cos (\alpha+\gamma) \leqslant \cos \left(60^{\circ}-(\alpha+\gamma)\right)+\cos (\alpha+\gamma) \Leftrightarrow \\ \cos (\alpha-\gamma) \leqslant \cos \left(60^{\circ}-(\alpha+\gamma)\right) \Leftrightarrow \\ \left|60^{\circ}-(\alpha+\gamma)\right| \leqslant 1 \alpha-\gamma+\left(\alpha \leqslant 30^{\circ} \text { and } \gamma \geqslant 30^{\circ}\right) \Leftrightarrow \\ 60^{\circ}-(\alpha+\gamma) \mid \leqslant \gamma-\alpha \end{array}$$

If
$$\begin{array}{c} 60^{\circ}-(\alpha+\gamma) \geqslant 0 \\ (3) 660^{\circ}-(\alpha+\gamma) \leqslant \gamma-\alpha \Leftrightarrow \gamma \geqslant 30^{\circ} \end{array}$$

If
$$\begin{array}{c} 60^{\circ}-(\alpha+\gamma) \leqslant 0 \\ (3) \Leftrightarrow(\alpha+\gamma)-60^{\circ} \leqslant \gamma-\alpha \Leftrightarrow \alpha \leqslant 30^{\circ} \end{array}$$

So inequality (3) holds.
$$\begin{array}{l} \cos \alpha \cos \beta \cos \gamma \leqslant \cos 30^{\circ} \cos \left(\alpha+\gamma-30^{\circ}\right) \cos \beta= \\ \cos 30^{\circ} \cos \left(90^{\circ}-\beta-30^{\circ}\right) \cos \beta= \\ \cos 30^{\circ} \cos \left(60^{\circ}-\beta\right) \cos \beta= \\ \frac{1}{2} \cos 30^{\circ}\left[\cos \left(60^{\circ}-2 \beta\right)+\cos 60^{\circ}\right] \leqslant \end{array}$$