(1) \(x^2-6x+9=0\)
(2) \(x^2+5x+3=0\)
(3) \(x^2-5x+4=0\)
(4) \(2x^2-6x+1=0\)
(5) \(4x^2-2x-1=0\)
(6) \(-2x^2-x-1=0\)
(7) \(5x^2-x-1=0\)
作者: admin
求証﹕\(\log_2(2^x+2^y)\geq\frac{x+y}{2}+1\).
証明﹕由AM-GM不等式得\(2^x+2^y\geq 2\sqrt{2^x\cdot 2^y}=2\sqrt{2^{x+y}}\)
\(\because y=\log_2x\)是增函數
\(\therefore \log_2(2^x+2^y)\geq \log_2{2\sqrt{2^{x+y}}}=\log_2 2+\log_2 2^{\frac{x+y}{2}}=1+\frac{x+y}{2}\)
証畢
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\(\frac{1372}{135}\)
\(x\in \mathbb{R}^+\),求下列各式的最小值.
(1) \(x^2+\frac{3}{x}\)
解﹕\(\because x>0\)
\(x^2+\frac{3}{x}=x^2+\frac{3}{2x}+\frac{3}{2x}\geq 3\sqrt[3]{x^2\cdot\frac{3}{2x}\cdot\frac{3}{2x}}=\frac{3}{2}\sqrt[3]{18}\)
當且僅當 \(x^2=\frac{3}{2x}\),即\(x=\frac{\sqrt[3]{12}}{2}\)時，\(x^2+\frac{3}{x}\)的最小值是 \(\frac{3}{2}\sqrt[3]{18}\)
(2) \(x+\frac{3}{x^2}\)
解﹕\(\because x>0\)
\(x+\frac{3}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{3}{x^2}\geq 3\sqrt[3]{\frac{x}{2}\cdot\frac{x}{2}\cdot\frac{3}{x^2}}=\frac{3}{2}\sqrt[3]{6}\)
當且僅當 \(\frac{x}{2}=\frac{3}{x^2}\),即\(x=\sqrt[3]{6}\)時，\(x+\frac{3}{x^2}\)的最小值是\(\frac{3}{2}\sqrt[3]{6}\)
求函數 \(2x+\frac{27}{x^2}\),\((x>0)\)的最小值
解﹕\(\because x>0\)
\(\therefore y=2x+\frac{27}{x^2}\)
\(=x+x+\frac{27}{x^2}\geq 3\sqrt[3]{x\cdot x\cdot \frac{27}{x^2}}=9\)
當且僅當 \(x=\frac{27}{x^2}\),即\(x=3\)時,\(y\)的最小值是\(9\)
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
解下列不等式
(1) x
例(1). 已知 \(x,y,z\) 為正數，求証﹕\(\frac{x}{y}+\frac{y}{x}\geq 2.\)
証﹕由均值不等式得 \(\frac{x}{y}+\frac{y}{x}\geq 2\sqrt{\frac{x}{y}\cdot\frac{y}{x}}=2\)
故原不等式成立。
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例(2). 已知 \(x,y,z\) 為正數，求証﹕\(\frac{x}{y}+\frac{y}{z}+\frac{x}{z}\geq 3.\)
証﹕由均值不等式得 \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq 3\sqrt[3]{\frac{x}{y}\cdot\frac{y}{z}\cdot\frac{z}{x}}=3\)
故原不等式成立。
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