回復 5# wen0623 的帖子
設 \(\overline{AP}=x\), \(\overline{AQ}=y\)
依題意即在 \(\displaystyle x+y=\frac{3a}{2}\) 限制下，求 \(\overline{PQ}=\sqrt{x^2+y^2-xy}\) 之最小值
由算幾不等式 \(\displaystyle \frac{x+y}{2}\geq\sqrt{xy}\) 可得 \(\displaystyle xy\leq \frac{9a^2}{16}\)
故 \(\displaystyle \sqrt{x^2+y^2-xy}=\sqrt{\frac{9a^2}{4}-3xy}\geq \sqrt{\frac{9a^2}{16}}=\frac{3a}{4}\)
等號成立時機：\(\displaystyle x=y=\frac{3a}{4}\)
得 \(\displaystyle \min \overline{PQ}=\frac{3a}{4}\)
