Problem 1677

41. To prove $S_{1} \geqslant \frac{2}{3} S_{2}$, it suffices to prove that $S_{\triangle A G H}+S_{\triangle A C H}+ S_{\triangle A G H} \leqslant \frac{1}{3} S_{2}$. Noting the parallelograms $A G O H, B I O D$, and $C E O F$, the solution to the proposition lies in proving
$$S_{\triangle O I F}+S_{\triangle O E H}+S_{\triangle O G D} \geqslant \frac{1}{3} S_{2}$$

Let $B C=a, C A=b, A B=c, I F=x, E H=y, G D=z$. Then equation (1) is equivalent to proving
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \geqslant \frac{1}{3}$$

According to the problem, we have $O E=C F$, thus $\frac{y}{b}=\frac{O E}{a}=\frac{C F}{a}$. Similarly, $\frac{z}{c}=\frac{B I}{a}$.
Therefore,
$$\frac{x}{a} \frac{y}{b} \frac{z}{c}=\frac{I F+C F+B I}{a}=1$$

Using equation (3), by the Cauchy-Schwarz inequality, we get $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \geqslant \frac{1}{3}\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^{2}=\frac{1}{3}$. Hence, equation (2) holds, and the proposition is proved.