利用廣義柯西不等式的
另解:
題目:已知 \(\displaystyle \Gamma_{-}: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 (a>0, b>0)\),過 \((3\sqrt{3},1)\),
則 \(a+b\) 之最小值為?此時,\(\Gamma_{-}\) 的方程式為?
解答:
因為 \(\displaystyle \Gamma_{-}: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\),過 \((3\sqrt{3},1)\),
帶入可得 \(\displaystyle \frac{27}{a^2}+\frac{1}{b^2}=1\),
由廣義柯西不等式,可得
\(\displaystyle \left(\left(\frac{3}{\sqrt[3]{a^2}}\right)^3+\left(\frac{1}{\sqrt[3]{b^2}}\right)^3\right) \left(\left(\sqrt[3]{a}\right)^3+\left(\sqrt[3]{b}\right)^3\right) \left(\left(\sqrt[3]{a}\right)^3+\left(\sqrt[3]{b}\right)^3\right)\)
\(\displaystyle\geq\left(\frac{3}{\sqrt[3]{a^2}}\cdot\sqrt[3]{a}\cdot\sqrt[3]{a}+\frac{1}{\sqrt[3]{b^2}}\cdot\sqrt[3]{b}\cdot\sqrt[3]{b}\right)^3\)
\(\displaystyle \Leftrightarrow \left(a+b\right)^2\geq 64\)
且因為 \(a>0,b>0\),所以 \(a+b\geq 8\)
且當等號成立時,若且唯若 \(\displaystyle \frac{3}{\sqrt[3]{a^2}}:\frac{1}{\sqrt[3]{b^2}}=\sqrt[3]{a}:\sqrt[3]{b}=\sqrt[3]{a}:\sqrt[3]{b}\Leftrightarrow a=3b\)
帶入 \(a+b=8\),可得 \(a=6,b=2\)
亦即,此時 \(\Gamma_{-}\) 的方程式為 \(\displaystyle \frac{x^2}{36}+\frac{y^2}{4}=1.\)
111.7.15補充
95台中一中,
https://math.pro/db/viewthread.php?tid=987&page=2#pid22591