Prove the following algebraic inequality first.
Let $x, y, z$ all be positive numbers. Then we have
$$\begin{array}{l} \frac{x y}{(x+z)(y+z)}+\frac{y z}{(y+x)(z+x)}+\frac{z x}{(z+y)(x+y)} \\ \geqslant \frac{3}{4}, \end{array}$$

with equality holding if and only if $x=y=z$.
In fact,
$$\begin{array}{l} \text { Equation (1) } \Leftrightarrow 4[x y(x+y)+y z(y+z)+z x(z+x)] \\ \geqslant 3(x+y)(y+z)(z+x) \\ \Leftrightarrow x y(x+y)+y z(y+z)+z x(z+x) \geqslant 6 x y z \\ \Leftrightarrow\left(\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}\right)^{2}+\left(\sqrt{\frac{y}{z}}-\sqrt{\frac{z}{y}}\right)^{2}+ \\ \left(\sqrt{\frac{z}{x}}-\sqrt{\frac{x}{z}}\right)^{2} \geqslant 0 . \end{array}$$
1. The above inequality is obviously true, hence Equation (1) holds, with equality if and only if $x=y=z$.
Now we prove the original inequality.
As shown in Figure 1, let
$\triangle A B C$ have sides
$$\begin{array}{l} B C=a, \\ C A=b, \\ A B=c . \end{array}$$

In $\triangle A E F$,
by the Angle Bisector Theorem, we know
$$\frac{F A^{\prime}}{A^{\prime} E}=\frac{F A}{A E}=\frac{b \cos A}{c \cos A}=\frac{b}{c} \text {. }$$

Similarly, $\frac{F B^{\prime}}{B^{\prime} D}=\frac{F B}{B D}=\frac{a}{c}$.
$$\text { Therefore, } \frac{S_{\triangle A^{\prime} F B^{\prime}}}{S_{\triangle D E F}}=\frac{a b}{(a+c)(b+c)} \text {. }$$

Similarly, $\frac{S_{\triangle B^{\prime} D C^{\prime}}}{S_{\triangle O E F}}=\frac{b c}{(a+b)(a+c)}$,
$$\frac{S_{\triangle C E A^{*}}}{S_{\triangle O E F}}=\frac{c a}{(b+c)(a+b)} \text {. }$$

Thus, S_{\triangle A^{\prime} F B^{\prime}}+S_{\triangle B^{\prime} ' D C^{\prime}}+S_{\triangle C^{\prime} E E^{\prime}}
$$=\left[\frac{a b}{(a+c)(b+c)}+\frac{b c}{(a+b)(a+c)}+\frac{c a}{(a+b)(b+c)}\right] \beta_{\triangle D E F} \text {. }$$

In inequality (1), let $x=a, y=b, z=c$, we get
$S_{\triangle H^{\prime} F H^{\prime}}+S_{\triangle A^{\prime} D C^{\prime}}+S_{\triangle C^{\prime} E A^{\prime}} \geqslant \frac{3}{4} S_{\triangle O E F}$.
Hence $S_{\triangle A^{\prime} B^{\prime} G^{*}}$
$$\begin{array}{l} =S_{\triangle D E F^{\prime}}-\left(S_{\triangle A^{\prime} F R^{\prime}}+S_{\triangle H^{\prime} D C^{\prime}}+S_{\triangle C^{\prime} A^{\prime}}\right) \\ \leqslant \frac{1}{4} S_{\triangle D E F^{-}} \end{array}$$

with equality holding if and only if $\triangle A B C$ is an equilateral triangle.