月份: 2021 年 1 月

設 \(a,b,c\in\mathbb{R}^+\)，\(a+b+c=1\)，求\((1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\)的最小值。

解﹕\(\because 1+\frac{1}{a}\geq 2\sqrt{1\cdot\frac{1}{a}}=\frac{2}{\sqrt{a}}\) (當且僅當\(a=1\)時等號成立)

同理可証\(1+\frac{1}{b}\geq\frac{2}{\sqrt{b}}\) (當且僅當\(b=1\)時等號成立)

同理可証\(1+\frac{1}{c}\geq\frac{2}{\sqrt{c}}\) (當且僅當\(c=1\)時等號成立)

\(\therefore (1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\geq\frac{2}{\sqrt{a}}\cdot\frac{2}{\sqrt{b}}\cdot\frac{2}{\sqrt{c}}=\frac{8}{\sqrt{abc}}\)

另一方面\(abc\leq (\frac{a+b+c}{3})^3=\frac{1}{27}\) (當且僅當\(a=b=c=\frac{1}{3}\)時等號成立)

\(\therefore \frac{8}{\sqrt{abc}}\geq\frac{8}{\sqrt{\frac{1}{27}}}=24\sqrt{3}\)

此方法的\(a,b,c\)的取值不一致，故此方法無效。

---

設 \(a,b,c\in\mathbb{R}^+\)，\(a+b+c=1\)，求\((1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\)的最小值。

解﹕\(\because 1+\frac{1}{a}=1+\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}\geq 4\sqrt[4]{1\cdot\frac{1}{3a}\cdot\frac{1}{3a}\cdot\frac{1}{3a}}=\frac{4}{\sqrt[4]{27a}}\) (當且僅當\(a=\frac{1}{3}\)時等號成立)

同理可証\(1+\frac{1}{b}\geq\frac{4}{\sqrt[4]{27b}}\) (當且僅當\(b=\frac{1}{3}\)時等號成立)

同理可証\(1+\frac{1}{c}\geq\frac{4}{\sqrt[4]{27c}}\) (當且僅當\(c=\frac{1}{3}\)時等號成立)

\(\therefore (1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\geq\frac{64}{\sqrt[4]{27^3(abc)^3}}\)

另一方面\(abc\leq (\frac{a+b+c}{3})^3=\frac{1}{27}\) (當且僅當\(a=b=c=\frac{1}{3}\)時等號成立)

\(\therefore \sqrt[4]{27^3(abc)^3}=1\)

\(\therefore (1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\geq 64\)

所以當\(a=b=c=\frac{1}{3}\)時，\((1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\)的最小值是64.

已知 \(x,y,z\in \mathbb{R}^+\)，且\(x+4y+3z=6\)，求\(x^2yz^3\)的最大值。

解﹕\(\because x+4y+3z\)

\(=\frac{x}{2}+\frac{x}{2}+4y+z+z+z\)

\(\geq 6\sqrt[6]{\frac{x}{2}\cdot\frac{x}{2}\cdot 4y\cdot z\cdot z\cdot z}=6\sqrt[6]{x^2 y z^3}\)

\(\therefore x+4y+3z\geq 6\sqrt[6]{x^2 y z^3}\)

\(\therefore 6\geq 6\sqrt[6]{x^2 y z^3}\)

\(\therefore x^2 y z^3\leq 1\)

當且僅當 \(\frac{x}{2}=4y=z=1\)，即\(x=2,y=\frac{1}{4},z=1\)時，\(x^2 y z^3\) 的最大值是 1

已知 \(0<x<\frac{2}{5}\)，求下列式子的最大值.

(1) \(x(2-5x)\)

解﹕\(\because 0<x<\frac{2}{5}\)

\(x(2-5x)=\frac{1}{5}\cdot 5x(2-5x)\leq\frac{1}{5}(\frac{5x+(2-5x)}{2})^2=\frac{1}{5}\)

當且僅當 \(5x=2-5x\)，即 \(x=\frac{1}{5}\) 時，\(x(2-5x)\)的最大值是\(\frac{1}{5}\)

---

(2) \(x^2(2-5x)\)

解﹕\(\because 0<x<\frac{2}{5}\)

\(x^2(2-5x)=\frac{4}{25}\cdot\frac{5x}{2}\cdot\frac{5x}{2}(2-5x)\leq\frac{4}{25}\left(\frac{\frac{5x}{2}+\frac{5x}{2}+(2-5x)}{3}\right)^3\)

\(=\frac{4}{25}\cdot\frac{8}{27}=\frac{32}{675}\)

當且僅當 \(\frac{5x}{2}=2-5x\)，即 \(x=\frac{4}{15}\) 時，\(x^2(2-5x)\)的最大值是\(\frac{32}{675}\)

---

(3) \(x(2-5x)^2\)

解﹕\(\because 0<x<\frac{2}{5}\)

\(x(2-5x)^2\)

\(=\frac{1}{10}\cdot 10x(2-5x)(2-5x)\)

\(\leq\frac{1}{10}\left(\frac{10+(2-5x)+(2-5x)}{3}\right)^3\)

\(\frac{1}{10}\cdot(\frac{14}{3})^3=\frac{1}{10}\cdot\frac{2744}{27}=\frac{1372}{135}\)

當且僅當 \(10x=2-5x\)，即\(x=\frac{2}{15}\)時，\(x(2-5x)^2\)的最大值是\(\frac{1372}{135}\)

求函數 \(y=x^2(12-5x)\),\((0<x<\frac{12}{5})\) 的最大值

解﹕\(\because 0<x<\frac{12}{5}, \therefore 12-5x>0 \)

\(\therefore y=x^2(12-5)\)

\(=\frac{4}{25}\cdot\frac{5x}{2}\cdot\frac{5x}{2}\cdot (12-5x)\leq \frac{4}{25}(\frac{\frac{5x}{2}+\frac{5x}{2}+12-5x}{3})^3=\frac{256}{25}\)

當且僅當\(\frac{5x}{2}=12-5x\),即 \(x=\frac{8}{5}\)時，\(y\)有最大值是\(\frac{256}{25}\)

#P13