2.2 基本不等式
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【基础过关系列】2022-2023学年高一数学上学期同步知识点剖析精品讲义(人教A版2019)
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基础知识
基本不等式
若\(a>0\) ,\(b>0\),则 \(a+b \geq 2 \sqrt{a b}\)(当且仅当\(a=b\)时,等号成立).
① \(\dfrac{a+b}{2}\)叫做正数\(a ,b\)的算术平均数, \(\sqrt{a b}\)叫做正数\(a ,b\)的几何平均数.
② 基本不等式的几何证明
![](https://img2023.cnblogs.com/blog/1943255/202305/1943255-20230516105007955-1105377991.png )
(当点\(D、O\)重合,即\(a=b\)时,取到等号)
③运用基本不等式求解最值时,牢记:一正,二定,三等.
一正指的是\(a>0 ,b>0\);二定指的是ab是个定值,三等指的是不等式中取到等号.
【例1】 求函数 \(y=x+\dfrac{1}{x}(x<0)\)的最值.
误解 \(x+\dfrac{1}{x} \geq 2 \sqrt{x \cdot \dfrac{1}{x}}=2\),故最小值是\(2\).
误解分析 误解中套用基本不等式,\(a=x\), \(b=\dfrac{1}{x}\),忽略了\(a>0,b>0\)的前提条件!
正解 \(∵x<0\) ,\(∴-x>0\) , \(-\dfrac{1}{x}>0\), \(\therefore-x+\left(-\dfrac{1}{x}\right) \geq 2 \sqrt{-x \cdot\left(-\dfrac{1}{x}\right)}=2\)(当\(x=-1\)取到等号)
\(\therefore x+\dfrac{1}{x}=-\left(-x-\dfrac{1}{x}\right) \leq-2\),故函数 \(y=x+\dfrac{1}{x}(x<0)\)的最大值为\(-2\),没有最小值.
【例2】 求函数 \(y=x+\dfrac{1}{x-1}(x>1)\)的最值.
误解 \(y=x+\dfrac{1}{x-1} \geq 2 \sqrt{x \cdot \dfrac{1}{x-1}}\)
误解分析 套用基本不等式\(a=x\), \(b=\dfrac{1}{x-1}\),满足\(a、b\)均为正数,但是最后求不出最值,因为 \(a b=x \cdot \dfrac{1}{x-1}\)不是一定值.
正解 \(y=x+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+1 \geq 2 \sqrt{(x-1) \cdot \dfrac{1}{x-1}}+1=3\).(当\(x=2\)时取到等号)
(通过凑项得到定值“ \((x-1) \cdot \dfrac{1}{x-1}=1\)”)
故函数 \(y=x+\dfrac{1}{x-1}(x>1)\)的最小值为\(2\),没有最大值.
【例3】 求函数 \(y=\dfrac{x^{2}+5}{\sqrt{x^{2}+4}}\)的最值.
误解 \(y=\dfrac{x^{2}+5}{\sqrt{x^{2}+4}}=\dfrac{x^{2}+4+1}{\sqrt{x^{2}+4}}=\sqrt{x^{2}+4}+\dfrac{1}{\sqrt{x^{2}+4}}\)\(\geq 2 \sqrt{\sqrt{x^{2}+4} \cdot \dfrac{1}{\sqrt{x^{2}+4}}}=2\),即最小值为\(2\).
误解分析 在误解中把 \(a=\sqrt{x^{2}+4}, b=\dfrac{1}{\sqrt{x^{2}+4}}\),,满足了“一正二定”,但忽略了能否取到等号?若\(a=b\),则\(\sqrt{x^{2}+4}=\dfrac{1}{\sqrt{x^{2}+4}} \Rightarrow \sqrt{x^{2}+4}=1 \Rightarrow x^{2}=-3\)显然方程无解,即不等式取不到等号,只能说明\(\sqrt{x^{2}+4}+\dfrac{1}{\sqrt{x^{2}+4}}>2\),那它有最小值么?(正解看下文_)
基本不等式及其变形
\(\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}} \leq \sqrt{a b} \leq \dfrac{a+b}{2} \leq \sqrt{\dfrac{a^{2}+b^{2}}{2}}\)(当且仅当\(a=b\)时等号成立)
(调和均值≤几何均值≤算术均值≤平方均值)
以上不等式把常见的二元关系(倒数和,乘积,和,平方和)联系起来,我们要清楚它们在求最值中的作用.
① \(a+b \geq 2 \sqrt{a b}\),积定求和;
② \(a b \leq\left(\dfrac{a+b}{2}\right)^{2}\),和定求积:
③ \(a^{2}+b^{2} \geq \dfrac{(a+b)^{2}}{2}\) (联系了\(a+b\)与平方和 \(a^2+b^2\))
④ \(a b \leq \dfrac{a^{2}+b^{2}}{2}\) (联系了\(ab\)与平方和 \(a^2+b^2\))
【例1】若\(a>0\),\(b>0\),\(ab=4\),求\(a+b\)的最小值.
解析 因为 \(a+b \geq 2 \sqrt{a b}=4\)(当\(a=b=2\)时取到等号),所以\(a+b\)的最小值是\(4\).
【例2】若\(a>0\),\(b>0\),\(a+b=4\),求\(ab\)的最大值.
解析 因为 \(a b \leq\left(\dfrac{a+b}{2}\right)^{2}=4\)(当\(a=b=2\)时取到等号),所以\(ab\)的最大值是\(4\).
【练1】 若\(a+b=2\),求 \(3^{a}+3^{b}\)的最小值.
解析 \(3^{a}+3^{b} \geq 2 \cdot \sqrt{3^{a} \cdot 3^{b}}=2 \cdot \sqrt{3^{a+b}}=2 \times 3=6\)(当 \(a=b=\sqrt{2}\)时取到等号).
【练2】 若\(a+2b=2\),求\(ab\)的最大值.
解析 因为 \(a b=\dfrac{1}{2} a \cdot 2 b \leq \dfrac{1}{2} \cdot\left(\dfrac{a+2 b}{2}\right)^{2}=\dfrac{1}{2}\)(当 \(a=1, b=\dfrac{1}{2}\)时取到等号),所以\(ab\)的最大值是 \(\dfrac{1}{2}\).
对勾函数
① 概念
形如 \(y=x+\dfrac{a}{x}(a>0)\)的函数.
② 图像
![](https://img2023.cnblogs.com/blog/1943255/202305/1943255-20230516105127524-238674232.png )
③ 性质
函数图像关于原点对称,
在第一象限中,当 \(0<x<\sqrt{a}\)时,函数递减,当 \(x>\sqrt{a}\)时,函数递增.
④ 与基本不等式的关系
由图很明显得知当\(x>0\)时, \(x=\sqrt{a}\)时取到最小值 \(y_{\min }=2 \sqrt{a}\),
其与基本不等式 \(x+\dfrac{a}{x} \geq 2 \sqrt{x \cdot \dfrac{a}{x}}=2 \sqrt{a}\) ( \(x=\sqrt{a}\)时取到最小值)是一致的.
【例】求函数 \(y=\dfrac{x^{2}+5}{\sqrt{x^{2}+4}}\)的最值.
解析\(y=\frac{x^{2}+5}{\sqrt{x^{2}+4}}=\frac{x^{2}+4+1}{\sqrt{x^{2}+4}}=\sqrt{x^{2}+4}+\frac{1}{\sqrt{x^{2}+4}}\),
令 \(t=\sqrt{x^{2}+4}\),则\(t≥2\),
因为对勾函数 \(y=t+\dfrac{1}{t}\)在\([2 ,+∞)\)上单调递增,
当\(t=2\)时,取得最小值 \(\dfrac{5}{2}\).
故 \(y=\dfrac{x^{2}+5}{\sqrt{x^{2}+4}}\)的最小值为\(\dfrac{5}{2}\),无最大值.
【练】求函数 \(y=x+\dfrac{4}{x}, \quad\left(\dfrac{1}{2} \leq x \leq 4\right)\)的最大值与最小值.
解析 函数 \(y=x+\dfrac{4}{x}\)是对勾函数,由其图象可知,当 \(x=\dfrac{1}{2}\)时取到最大值 \(\dfrac{17}{2}\),当\(x=2\)时取到最小值\(4\).
基本方法
基本不等式常见的解题方法
方法1 直接法
【典题1】 下列命题正确的是( )
A.函数 \(y=x+\dfrac{1}{x}\)最小值为\(2\)
B.若\(a ,b∈R\)且\(ab>0\),则 \(\dfrac{b}{a}+\dfrac{a}{b} \geq 2\)
C.函数\(\sqrt{x^{2}+2}+\dfrac{1}{\sqrt{x^{2}+2}}\)的最小值为\(2\)
D.函数\(y=2-3 x-\dfrac{4}{x}\)的最小值为 \(2-4 \sqrt{3}\)
解析 \(A\)错误,当\(x<0\)时或\(≠1\)时不成立;\(B\)正确,
因为\(ab>0\),所以 \(\dfrac{b}{a}>0, \dfrac{a}{b}>0\),且 \(\dfrac{b}{a}+\dfrac{a}{b} \geq 2\);\(C\)错误,
若运用基本不等式,需 \(\sqrt{x^{2}+2^{2}}=1, \quad x^{2}=-1\)无实数解;\(D\)错误,
\(y=2-\left(3 x+\dfrac{4}{x}\right) \leq 2-4 \sqrt{3}\)
答案:\(B\)
点拨 注意理解使用基本不等式 \(a+b \geq 2 \sqrt{a b}\)的六字真言“一正二定三等”.
巩固练习
1.已知\(x>0 ,y>0\),且 \(\dfrac{1}{x}+\dfrac{9}{y}=1\),则\(xy\)的最小值为\(\underline{\quad \quad}\).
2.下列命题中正确的是( )
A.若\(a,b∈R\),则 \(\dfrac{b}{a}+\dfrac{a}{b} \geq 2 \sqrt{\dfrac{b}{a} \cdot \dfrac{a}{b}}=2\)
B.若\(x>0\),则 \(x+\dfrac{1}{x}>2\)
C.若\(x<0\),则 \(x+\dfrac{4}{x} \geq-2 \sqrt{x \cdot \dfrac{4}{x}}=-4\)
D.若\(x∈R\),则 \(2^{x}+2^{-x} \geq 2 \sqrt{2^{x} \cdot 2^{-x}}=2\)
参考答案
- 答案 \(36\)
解析 \(∵x>0 ,y>0\),且 \(\dfrac{1}{x}+\dfrac{9}{y}=1\),
由基本不等式可得 \(1 \geq 2 \sqrt{\dfrac{9}{x y}}\),当且仅当 \(\dfrac{1}{x}=\dfrac{9}{y}=\dfrac{1}{2}\)即\(x=2,y=18\)时取等号,
解可得\(xy≥36\),即\(xy\)的最小值\(36\). - 答案 \(D\)
解析 \(A\)选项必须保证\(a,b\),同号.\(B\)选项应取到等号,若\(x>0\),则\(x+\dfrac{1}{x} \geq 2\),\(C\)选项应该为\(≤\),故选:\(D\).
方法2 凑项法
【典题1】 函数 \(y=2 x+\dfrac{2}{x-1}(x>1)\)的最小值是( )
A.\(2\) \(\qquad \qquad\) B.\(4\) \(\qquad \qquad\) C.\(6\) \(\qquad \qquad\) D.\(8\)
解析 因为 \(y=2 x+\dfrac{2}{x-1}(x>1)=2(x-1)+\dfrac{2}{x-1}+2\)\(\geq 2 \sqrt{2(x-1) \cdot \dfrac{2}{x-1}}+2=6\),
当且仅当 \(2(x-1)=\dfrac{2}{x-1}\)即\(x=2\)时取等号,此时取得最小值\(6\).
故选:\(C\).
点拨 本题不能直接使用基本不等式,因为 \(2 x \times \dfrac{2}{x-1}\)不是定值,故通过凑项,得到 \(2(x-1) \cdot \dfrac{2}{x-1}=4\)定值.
巩固练习
1.若\(x>0\),则函数 \(y=x+\dfrac{1}{2 x+1}\)的最小值为( )
A.\(\sqrt{2}+\dfrac{1}{2}\) \(\qquad \qquad\) B.\(\sqrt{2}-\dfrac{1}{2}\) \(\qquad \qquad\) C.\(\sqrt{2}+1\) \(\qquad \qquad\) D.\(\sqrt{2}-1\)
2.若\(a ,b>0\),\(ab+2a+b=4\),则\(a+b\)的最小值为( )
A.\(2\) \(\qquad \qquad\) B. \(\sqrt{6}-1\) \(\qquad \qquad\) C. \(2 \sqrt{6}-2\) \(\qquad \qquad\) D. \(2 \sqrt{6}-3\)
参考答案
- 答案 \(B\)
解析 \(x>0\),函数 \(y=x+\dfrac{1}{2 x+1}=\left(x+\dfrac{1}{2}\right)+\left(\dfrac{\dfrac{1}{2}}{x+\dfrac{1}{2}}\right)-\dfrac{1}{2} \geq 2 \sqrt{\dfrac{1}{2}}-\dfrac{1}{2}=\sqrt{2}-\dfrac{1}{2}\),
当且仅当 \(x=\dfrac{\sqrt{2}-1}{2}\)时取等号.
\(∴\)函数 \(y=x+\dfrac{1}{2 x+1}\)的最小值为 \(\sqrt{2}-\dfrac{1}{2}\).故选:\(B\). - 答案 \(D\)
解析 \(∵a ,b=R^*\) ,\(ab+2a+b=4\),\(∴b(a+1)=4-2a\),
\(\therefore b=\dfrac{4-2 a}{a+1}=-\dfrac{2 a-4}{a+1}=-\dfrac{2(a+1)-6}{a+1}=-2+\dfrac{6}{a+1}\),
\(\therefore a+b=a-2+\dfrac{6}{a+1}=a+1+\dfrac{6}{a+1}-3\)
\(∵a>0 ,b>0\), \(\therefore a+b \geq 2 \sqrt{(a+1) \cdot \dfrac{6}{a+1}}-3=2 \sqrt{6}-3\)
当且仅当 \(a+1=\dfrac{6}{a+1}\)即 \(a=\sqrt{6}-1\)时″\(=\)″,故选:\(D\).
方法3 凑系数法
【典题1】 当\(0<x<4\)时,则\(y=x(8-2x)\)的最大值为\(\underline{\quad \quad}\).
解析 \(y=x(8-2 x)=\dfrac{1}{2}[2 x \cdot(8-2 x)] \leq \dfrac{1}{2}\left(\dfrac{2 x+8-2 x}{2}\right)^{2}=8\)
当\(2x=8-2x\),即\(x=2\)时取等号 当\(x=2\)时,\(y=x(8-2x)\)的最大值为\(8\).
点拨 ① \(a+b \geq 2 \sqrt{a b}\),积定求和;② \(a b \leq\left(\dfrac{a+b}{2}\right)^{2}\),和定求积;
本题使用不等式②,为了使得\(a+b\)是定值,需要凑系数,使得\(2x+8-2x=2\)为定值.当然本题也可以用二次函数求最值.
巩固练习
1.设 \(0<x<\dfrac{3}{2}\),则函数\(y=4x(3-2x)\)的最大值为\(\underline{\quad \quad}\).
2.已知\(a ,b\)为正数, \(4a^2+b^2=7\),则 \(a \sqrt{1+b^{2}}\)的最大值为\(\underline{\quad \quad}\).
参考答案
- 答案 \(A\)
解析 \(\because 0<x<\dfrac{3}{2}\),\(∴3-2x>0\),
\(\therefore y=4 x(3-2 x)=2 \cdot 2 x(3-2 x) \leq 2\left(\dfrac{2 x+3-2 x}{2}\right)^{2}=\dfrac{9}{2}\),
当且仅当\(2x=3-2x\)即 \(x=\dfrac{3}{4} \in\left(0, \dfrac{3}{2}\right)\)时等号成立. - 答案 \(2\)
解析 因为 \(4a^2+b^2=7\),
则 \(a \sqrt{1+b^{2}}=\dfrac{1}{2}(2 a) \sqrt{1+b^{2}}=\dfrac{1}{2} \sqrt{4 a^{2}\left(1+b^{2}\right)}\)\(\leq \dfrac{1}{2} \times \dfrac{4 a^{2}+1+b^{2}}{2}=2\),
当且仅当 \(4a^2=1+b^2\)时,取得最大值.
方法4 巧“1”法
【典题1】 若正数\(x ,y\)满足 \(\dfrac{3}{x}+\dfrac{1}{y}=5\),则\(3x+4y\)的最小值是\(\underline{\quad \quad}\).
解析 \(\because \dfrac{3}{x}+\dfrac{1}{y}=5\), \(\therefore \dfrac{1}{5}(3 x+4 y)=1\),
\(\therefore 3 x+4 y=(3 x+4 y) \times 1=\dfrac{1}{5}(3 x+4 y)\left(\dfrac{3}{x}+\dfrac{1}{y}\right)\)
\(=\dfrac{1}{5}\left(9+4+\dfrac{12 y}{x}+\dfrac{3 x}{y}\right) \geq \dfrac{1}{5}(13+2 \sqrt{36})=5\),当且仅当 \(\dfrac{12 y}{x}=\dfrac{3 x}{y}\)时等号成立.
点拨 本题巧妙得利用\(3x+4y=(3x+4y)×1\)得到" \(\dfrac{12 y}{x}+\dfrac{3 x}{y}\)"这符合使用基本不等式的“模型”.
巩固练习
1.已知\(a>0 ,b>0\)且\(a+b=1\),则 \(\dfrac{1}{a}+\dfrac{2}{b}\)的最小值为\(\underline{\quad \quad}\).
2.若\(a>0 ,b>0\)且\(a+b=4\),则下列不等式恒成立的是 ( )
A. \(\dfrac{1}{a b} \leq \dfrac{1}{4}\) \(\qquad \qquad\) B. \(\dfrac{1}{a}+\dfrac{1}{b} \leq 1\) \(\qquad \qquad\) C. \(\sqrt{a b} \geq 2\) \(\qquad \qquad\) D. \(a^2+b^2≥8\)
参考答案
- 答案 \(3+2 \sqrt{2}\)
解析 \(∵a+b=1\),
\(\therefore \dfrac{1}{a}+\dfrac{2}{b}=(a+b)\left(\dfrac{1}{a}+\dfrac{2}{b}\right)=3+\dfrac{b}{a}+\dfrac{2 a}{b}\)\(\geq 3+2 \sqrt{\dfrac{b}{a} \cdot \dfrac{2 a}{b}}=3+2 \sqrt{2}\),
当且仅当 \(\dfrac{b}{a}=\dfrac{a}{b}\),即 \(a=b=\dfrac{1}{2}\)时,取等号. - 答案 \(D\)
解析 方法一:取特殊值排除法
令\(a=1 ,b=3\),经过检验\(ABC\)项都错,\(D\)对,故选\(D\).
方法二:\(∵a+b=4\), \(\therefore 4=a+b \geq 2 \sqrt{a b} \Rightarrow \sqrt{a b} \leq 2\),所以\(C\)错;
\(\sqrt{a b} \leq 2 \Rightarrow a b \leq 4 \Rightarrow \dfrac{1}{a b} \geq \dfrac{1}{4}\),所以\(A\)错;
\(\dfrac{1}{a}+\dfrac{1}{b}=\left(\dfrac{1}{a}+\dfrac{1}{b}\right) \times 1=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{a+b}{4}\right)\)\(=\dfrac{1}{4}\left(2+\dfrac{b}{a}+\dfrac{a}{b}\right) \geq \dfrac{1}{4}(2+2)=1\),所以\(B\)错;
\(a^2+b^2≥2ab≥8\),所以\(D\)对.
方法5 换元法
【典题1】 已知\(x>2\),则函数 \(y=\dfrac{x^{2}-4 x+8}{x-2}\)的最小值是\(\underline{\quad \quad}\).
解析 \(y=\dfrac{x^{2}-4 x+8}{x-2}=\dfrac{(x-2)^{2}+4}{x-2}=(x-2)+\dfrac{4}{x-2}\),
\(∵x>2 ,∴x-2>0\),
\(\therefore y=(x-2)+\dfrac{4}{x-2} \geq 2 \sqrt{(x-2) \dfrac{4}{x-2}}=2 \sqrt{4}=4\),
当且仅当 \(x-2=\dfrac{4}{x-2}\),即\(x=4\)时取等号,
故最小值为\(4\).
点拨 该题型属于形如 \(y=\dfrac{a x^{2}+b x+c}{d x^{2}+e x+f}\)的最值问题,常用换元法与基本不等式处理.
巩固练习
1.若\(x>1\),则 \(y=\dfrac{x-1}{x^{2}+x-1}\)的最大值为( )
A. \(\dfrac{1}{6}\) \(\qquad \qquad\) B. \(\dfrac{1}{4}\) \(\qquad \qquad\) C. \(\dfrac{1}{5}\) \(\qquad \qquad\) D. \(\dfrac{1}{3}\)
2.若\(a ,b∈R^*\),\(a+b=1\),求 \(\sqrt{a+\dfrac{1}{2}}+\sqrt{b+\dfrac{1}{2}}\)的最大值.
参考答案
- 答案 \(C\)
解析 令\(t=x-1\),则\(x=t+1\),\(t>0\),
原式 \(=\dfrac{t}{(t+1)^{2}+(t+1)-1}=\dfrac{t}{t^{2}+3 t+1}=\dfrac{1}{t+\dfrac{1}{t}+3}\)\(\leq \dfrac{1}{\sqrt[2]{t \cdot \dfrac{1}{t}+3}}=\dfrac{1}{5}\),
当且仅当\(t=1\)即\(x=2\)时等号成立,
故选:\(C\). - 答案 \(2\)
解析 设 \(s=\sqrt{a+\dfrac{1}{2}}\) , \(t=\sqrt{b+\dfrac{1}{2}}\),则 \(a=s^{2}-\dfrac{1}{2}\) , \(b=t^{2}-\dfrac{1}{2}\),
\(∵a+b=1\) \(\therefore s^{2}+t^{2}=2\)
\(\therefore \dfrac{s+t}{2} \leq \sqrt{\dfrac{s^{2}+t^{2}}{2}}=1 \Rightarrow s+t \leq 2\),即 \(\sqrt{a+\dfrac{1}{2}}+\sqrt{b+\dfrac{1}{2}} \leq 2\).
分层练习
【A组---基础题】
1.已知\(a ,b\)为实数,且\(a⋅b≠0\),则下列命题错误的是 ( )
A.若\(a>0 ,b>0\),则 \(\dfrac{a+b}{2} \geq \sqrt{a b}\)
B.若 \(\dfrac{a+b}{2} \geq \sqrt{a b}\),则\(a>0 ,b>0\)
C.若\(a≠b\),则 \(\dfrac{a+b}{2}>\sqrt{a b}\)
D.若 \(\dfrac{a+b}{2}>\sqrt{a b}\),则\(a≠b\)
2.已知\(a≥0\) ,\(b≥0\),且\(a+b=2\),则( )
A. \(a b \leq \dfrac{1}{2}\) \(\qquad \qquad\) B. \(a b \geq \dfrac{1}{2}\) \(\qquad \qquad\) C. \(a^2+b^2≥2\) \(\qquad \qquad\) D. \(a^2+b^2≤3\)
3.(多选)设\(a>0\),\(b>0\),且\(a+2b=4\),则下列结论正确的是( )
A. \(\dfrac{1}{a}+\dfrac{1}{b}\)的最小值为 \(\sqrt{2}\)
B. \(\dfrac{2}{a}+\dfrac{1}{b}\)的最小值为\(2\)
C. \(\dfrac{1}{a}+\dfrac{2}{b}\)的最小值为 \(\dfrac{9}{4}\)
D. \(\dfrac{b}{a+1}+\dfrac{a}{b+1}>\dfrac{8}{7}\)恒成立
4.已知\(a ,b∈R\),如果\(ab=1\),那么\(a+b\)的最小值为\(\underline{\quad \quad}\);如果\(a+b=1\),那么\(ab\)的最大值为\(\underline{\quad \quad}\).
5.若实数\(a, b\) 满足 \(\dfrac{1}{a}+\dfrac{2}{b}=\sqrt{a b}\),则\(ab\)的最小值为\(\underline{\quad \quad}\).
6.已知\(x,y∈R^+\),若\(x+y+xy=8\),则\(xy\)的最大值为\(\underline{\quad \quad}\).
7.已知\(x>2\),则 \(y=x+\dfrac{1}{x-2}\)的最小值是\(\underline{\quad \quad}\).
8.若\(0<2x<3\),则\((3-2x)x\)的最大值为\(\underline{\quad \quad}\) .
9.若正实数\(a ,b\),满足\(a+b=1\),则 \(\dfrac{b}{3 a}+\dfrac{3}{b}\)的最小值为 .
10.设\(x>-1\),求 \(y=\dfrac{(x+5)(x+2)}{x+1}\)的最小值.
参考答案
- 答案 \(C\)
- 答案 \(C\)
解析 由\(a≥0\) ,\(b≥0\),且\(a+b=2\)
\(∴4=(a+b)^2=a^2+b^2+2ab≤2(a^2+b^2 )\),当且仅当\(a=b=1\)时等号成立
\(∴a^2+b^2≥2\). - 答案 \(BC\)
解析 因为\(a>0,b>0\),且\(a+2b=4\),
对于\(A\), \(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)(a+2 b)=\dfrac{1}{4}\left(3+\dfrac{2 b}{a}+\dfrac{a}{b}\right) \geq \dfrac{1}{4}(3+2 \sqrt{2})\),
当且仅当 \(a=4 \sqrt{2}-4, b=4-2 \sqrt{2}\)时取等号,故选项\(A\)错误;
对于\(B\), \(\dfrac{2}{a}+\dfrac{1}{b}=\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)(a+2 b)=\dfrac{1}{4}\left(4+\dfrac{4 b}{a}+\dfrac{a}{b}\right) \geq \dfrac{1}{4}(4+4)=2\),
当且仅当\(a=2 ,b=1\)时取等号,故选项\(B\)正确;
对于\(C\), \(\dfrac{1}{a}+\dfrac{2}{b}=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)(a+2 b)=\dfrac{1}{4}\left(5+\dfrac{2 b}{a}+\dfrac{2 a}{b}\right)=\dfrac{1}{4}(5+4)=\dfrac{9}{4}\),
当且仅当\(a=\dfrac{4}{3}, b=\dfrac{4}{3}\)时取等号,故选项\(C\)正确;
对于\(D\),当 \(a=\dfrac{4}{3}, b=\dfrac{4}{3}\)时,\(a+2b=4\),但 \(\dfrac{b}{a+1}+\dfrac{a}{b+1}=\dfrac{\dfrac{4}{3}}{\dfrac{4}{3}+1}+\dfrac{\dfrac{4}{3}}{\dfrac{4}{3}+1}=\dfrac{8}{7}\),故选项\(D\)错误.
故选:\(BC\). - 答案 \(2, \dfrac{1}{4}\)
解析 因为\(a ,b∈R\),所以 \(\dfrac{a+b}{2} \geq \sqrt{a b}\),
所以 \(a+b \geq 2 \sqrt{a b}=2\).
故当\(ab=1\)时,\(a+b\)取最小值\(2\),此时\(a=b=1\).
又当\(a+b=1\)时, \(\sqrt{a b} \leq \dfrac{a+b}{2}=\dfrac{1}{2}\).所以 \(a b \leq \dfrac{1}{4}\). - 答案 \(2 \sqrt{2}\)
解析 由题易知\(a>0 ,b>0\),
\(\because \dfrac{1}{a}+\dfrac{2}{b} \geq 2 \sqrt{\dfrac{2}{a b}}\),当 \(\dfrac{1}{a}=\dfrac{2}{b}\),即\(b=2a\)时取到等号,
\(\therefore \sqrt{a b} \geq 2 \sqrt{\dfrac{2}{a b}} \Rightarrow a b \geq 2 \sqrt{2}\).
由 \(\left\{\begin{array}{c} b=2 a \\ \dfrac{1}{a}+\dfrac{2}{b}=\sqrt{a b} \end{array}\right.\),解得 \(a=2^{\dfrac{1}{4}}, b=2^{\dfrac{5}{4}}\),
即当 \(a=2^{\dfrac{1}{4}}, b=2^{\dfrac{5}{4}}\)时,\(ab\)取到最小值 \(2 \sqrt{2}\). - 答案 \(4\)
解析 \(∵\)正数\(x,y\)满足\(x+y+xy=8\),
\(\therefore 8-x y=x+y \geq 2 \sqrt{x y}\), \(x y+2 \sqrt{x y}-8 \leq 0\),
解得 \(0<\sqrt{x y} \leq 2\),
故\(xy≤4\),当且仅当\(x=y=2\)时取等号.
\(∴xy\)的最大值为\(4\). - 答案 \(4\)
解析 \(∵x>2 ,∴x-2>0\),
\(\therefore y=x+\dfrac{1}{x-2}=x-2+\dfrac{1}{x-2}+2 \geq 2+2=4\)(当 \(x-2=\dfrac{1}{x-2}\),即\(x=3\)时取到等号).
\(\therefore y=x+\dfrac{1}{x-2}\)的最小值是\(4\). - 答案 \(\dfrac{9}{8}\)
解析 \(∵0<2x<3\),\(∴3-2x>0,x>0\),
\(\therefore(3-2 x) x=\dfrac{1}{2}(3-2 x) \cdot 2 x \leq \dfrac{1}{2}\left(\dfrac{3-2 x+2 x}{2}\right)^{2}=\dfrac{9}{8}\),
当且仅当\(3-2x=2x\),即 \(x=\dfrac{3}{4}\)时取等号,
\(∴(3-2x)x\)的最大值为 \(\dfrac{9}{8}\). - 答案 \(5\)
解析 根据题意,若正实数\(a,b\),满足\(a+b=1\),
则 \(\dfrac{b}{3 a}+\dfrac{3}{b}=\dfrac{b}{3 a}+\dfrac{3 a+3 b}{b}=\dfrac{b}{3 a}+\dfrac{3 a}{b}+3\)\(\geq 2 \times \sqrt{\dfrac{b}{3 a} \times \dfrac{3 a}{b}}+3=5\),
当且仅当 \(b=3 a=\dfrac{3}{4}\)时等号成立,
即 \(\dfrac{b}{3 a}+\dfrac{3}{b}\)的最小值为\(5\). - 答案 \(9\)
解析, 设\(x+1=t>0\),则\(x=t-1\),
于是有\(y=\dfrac{(t+4)(t+1)}{t}=\dfrac{t^{2}+5 t+4}{t}=t+\dfrac{4}{t}+5 \geq 2 \sqrt{t \cdot \dfrac{4}{t}}+5=9\)
当且仅当 \(t=\dfrac{4}{t}\),即\(t=2\)时取等号,此时\(x=1\).
\(∴\)当\(x=1\)时,函数取得最小值是\(9\).
【B组---提高题】
1.(多选)下列说法正确的是( )
A. \(x+\dfrac{1}{x}(x>0)\)的最小值是\(2\)
B. \(\dfrac{x^{2}+2}{\sqrt{x^{2}+2}}\)的最小值是 \(\sqrt{2}\)
C. \(\dfrac{x^{2}+5}{\sqrt{x^{2}+4}}\)的最小值是\(2\)
D. \(2-3 x-\dfrac{4}{x}\)的最大值是 \(2-4 \sqrt{3}\)
2.若实数\(m,n>0\),满足\(2m+n=1\),以下选项中正确的有( )
A.\(mn\)的最小值为 \(\dfrac{1}{8}\)
B. \(\dfrac{1}{m}+\dfrac{1}{n}\)的最小值为 \(4 \sqrt{2}\)
C. \(\dfrac{2}{m+1}+\dfrac{9}{n+2}\)的最小值为\(5\)
D. \(4m^2+n^2\)的最小值为 \(\dfrac{1}{2}\)
3.当\(x >1\)时,不等式 \(x+\dfrac{1}{x-1} \geq a\)恒成立,则实数a的最大值为\(\underline{\quad \quad}\).
4.已知\(ab>0\),\(a+b=5\),则 \(\dfrac{2}{a+1}+\dfrac{1}{b+1}\)的最小值为\(\underline{\quad \quad}\).
5.若实数\(x ,y\)满足 \(x^2+y^2+xy=1\),则\(x+y\)的最大值\(\underline{\quad \quad}\) .
6.已知正实数\(a,b\)满足\(a+b=1\),则 \(\dfrac{a^{2}+4}{a}+\dfrac{b^{2}+1}{b}\)的最小值为\(\underline{\quad \quad}\).
7.已知正实数\(x,y\)满足\(x+y=1\),则 \(\dfrac{y}{x}+\dfrac{2}{x y}\)的最小值为\(\underline{\quad \quad}\).
8.已知正实数\(a ,b\)满足:\(a+b=1\),则 \(\dfrac{2 a}{a^{2}+b}+\dfrac{b}{a+b^{2}}\)的最大值是\(\underline{\quad \quad}\) .
参考答案
- 答案 \(AB\)
解析 由基本不等式可知,\(x>0\)时, \(x+\dfrac{1}{x} \geq 2\),当且仅当 \(x=\dfrac{1}{x}\)即\(x=1\)时取等号,故\(A\)正确;
\(B\): \(\dfrac{x^{2}+2}{\sqrt{x^{2}+2}}=\sqrt{x^{2}+2} \geq \sqrt{2}\),当\(x=0\)时取得等号,故\(B\)正确;
\(C\): \(\dfrac{x^{2}+5}{\sqrt{x^{2}+4}}=\sqrt{x^{2}+4}+\dfrac{1}{\sqrt{x^{2}+4}}\),令 \(t=\sqrt{x^{2}+4}\),则\(t≥2\),
因为 \(y=t+\dfrac{1}{t}\)在\([2 ,+∞)\)上单调递增,当\(t=2\)时,取得最小值 \(\dfrac{5}{2}\),故\(C\)错误;
\(D\): \(2-\left(3 x+\dfrac{4}{x}\right)\)在\(x<0\)时,没有最大值,故\(D\)错误.
故选:\(AB\). - 答案 \(D\)
解析 \(∵\)实数\(m,n>0\), \(\therefore 2 m+n=1 \geq 2 \sqrt{2 m n}\),
整理得 \(m n \leq \dfrac{1}{8}\),当且仅当 \(\left\{\begin{array}{l} n=\dfrac{1}{2} \\ m=\dfrac{1}{4} \end{array}\right.\)时取“\(=\)“,故选项\(A\)错误;
\(\because \dfrac{1}{m}+\dfrac{1}{n}=(2 m+n)\left(\dfrac{1}{m}+\dfrac{1}{n}\right)=3+\dfrac{n}{m}+\dfrac{2 m}{n} \geq 3+2 \sqrt{2}\),
当且仅当 \(\left\{\begin{array}{l} m=\dfrac{2-\sqrt{2}}{2} \\ n=\sqrt{2}-1 \end{array}\right.\)时取“\(=\)“,故选项\(B\)错误;
\(∵2m+n=1\),\(∴2(m+1)+(n+2)=5\),
\(\therefore \dfrac{2}{m+1}+\dfrac{9}{n+2}=\dfrac{1}{5}[2(m+1)+(n+2)]\left(\dfrac{2}{m+1}+\dfrac{9}{n+2}\right)\)
\(=\dfrac{1}{5}\left[13+\dfrac{2(n+2)}{m+1}+\dfrac{18(m+1)}{n+2}\right] \geq \dfrac{1}{5}(13+2 \sqrt{36})=5\),当且仅当 \(\left\{\begin{array}{l} m=0 \\ n=1 \end{array}\right.\)时取“\(=\)“,
\(\therefore \dfrac{2}{m+1}+\dfrac{9}{n+2}>5\),故选项\(C\)错误;
\(∵2m+n=1\),
\(\therefore 1=(2 m+n)^{2}=4 m^{2}+n^{2}+4 m n=4 m^{2}+n^{2}+2 \sqrt{4 m^{2}} \cdot \sqrt{n^{2}} \leq 2\left(4 m^{2}+n^{2}\right)\),
\(\therefore 4 m^{2}+n^{2} \geq \dfrac{1}{2}\),当且仅当 \(\left\{\begin{array}{l} n=\dfrac{1}{2} \\ m=\dfrac{1}{4} \end{array}\right.\)时取“\(=\)“,故选项\(D\)正确,
故选:\(D\). - 答案 \(3\)
解析 \(x+\dfrac{1}{x-1} \geq a\)恒成立 \(\Leftrightarrow\left(x+\dfrac{1}{x-1}\right)_{\min } \geq a\),
因为\(x >1\),即\(x-1 >0\),
所以 \(x+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+1 \geq 2 \sqrt{(x-1) \cdot \dfrac{1}{x-1}}+1=3\),
当且仅当 \(x-1=\dfrac{1}{x-1}\),即\(x=2\)时,等号成立.
所以\(a≤3\),即\(a\)的最大值为\(3\). - 答案 \(\dfrac{3+2 \sqrt{2}}{7}\)
解析 \(∵ab>0,a+b=5\)知\(a>0 ,b>0\),
又\(a+1+b+1=7\), \(\therefore \dfrac{1}{7}(a+1+b+1)=1\),
而 \(\dfrac{2}{a+1}+\dfrac{1}{b+1}=\dfrac{1}{7}(a+1+b+1)\left(\dfrac{2}{a+1}+\dfrac{1}{b+1}\right)\)\(=\dfrac{1}{7}\left(3+\dfrac{2(b+1)}{a+1}+\dfrac{a+1}{b+1}\right) \geq \dfrac{1}{7}(3+2 \sqrt{2})\),
经检验等号成立,故填 \(\dfrac{3+2 \sqrt{2}}{7}\). - 答案 \(\dfrac{2 \sqrt{3}}{3}\)
解析 \(\because(x+y)^{2}=x^{2}+y^{2}+2 x y=1-x y+2 x y=1+x y\),
\(\therefore(x+y)^{2}-1=x y\),\(\because x y \leq \dfrac{(x+y)^{2}}{4}\),
\(\therefore(x+y)^{2}-1 \leq \dfrac{(x+y)^{2}}{4}\),解得 \((x+y)^{2} \leq \dfrac{4}{3}\),
\(\therefore-\dfrac{2 \sqrt{3}}{3} \leq x+y \leq \dfrac{2 \sqrt{3}}{3}\),则\(x+y\)的最大值为 \(\dfrac{2 \sqrt{3}}{3}\). - 答案 \(10\)
解析 由 \(\dfrac{a^{2}+4}{a}+\dfrac{b^{2}+1}{b}=a+b+\dfrac{4}{a}+\dfrac{1}{b}=1+\dfrac{4}{a}+\dfrac{1}{b}\)
\(∵a+b=1\),
\(\therefore \dfrac{4}{a}+\dfrac{1}{b}=\left(\dfrac{4}{a}+\dfrac{1}{b}\right)(a+b)=5+\dfrac{4 b}{a}+\dfrac{a}{b} \geq 2 \sqrt{\dfrac{4 b}{a} \times \dfrac{a}{b}}+5=9\),
当且仅当 \(b=\dfrac{1}{3}, \quad a=\dfrac{2}{3}\)时取等号.
\(\therefore \dfrac{a^{2}+4}{a}+\dfrac{b^{2}+1}{b}\)的最小值为\(9+1=10\). - 答案 \(4+2 \sqrt{6}\)
解析 \(∵\)正实数\(x ,y\)满足\(x+y=1\),
\(∴y=1-x,x∈(0 ,1)\),
\(\therefore \dfrac{y}{x}+\dfrac{2}{x y}=\dfrac{1-x}{x}+\dfrac{2}{x(1-x)}=-1+\dfrac{1}{x}+\dfrac{2}{x(1-x)}=-1+\dfrac{3-x}{x(1-x)}\),
令\(t=3-x∈(2 ,3)\),
则 \(\dfrac{y}{x}+\dfrac{2}{x y}=-1+\dfrac{t}{(3-t)(t-2)}=-1+\dfrac{t}{-t^{2}-6+5 t}=-1+\dfrac{1}{5-\left(t+\dfrac{6}{t}\right)}\)\(\geq-1+\dfrac{1}{5-2 \sqrt{t \cdot \dfrac{6}{t}}}=-1+5+2 \sqrt{6}=4+2 \sqrt{6}\),
当且仅当 \(t=\sqrt{6}\)时取“\(=\)”. - 答案 \(\dfrac{2 \sqrt{3}+3}{3}\)
解析 \(\dfrac{2 a}{a^{2}+b}+\dfrac{b}{a+b^{2}}=\dfrac{2 a}{a^{2}+1-a}+\dfrac{1-a}{a+(1-a)^{2}}=\dfrac{a+1}{a^{2}-a+1}\),
由题意得,\(0<a<1\),令\(a+1=t∈(1 ,2)\),
\(\therefore \dfrac{a+1}{a^{2}-a+1}=\dfrac{t}{(t-1)^{2}-(t-1)+1}=\dfrac{1}{t+\dfrac{3}{t}-3}\)\(\leq \dfrac{1}{2 \sqrt{3}-3}=\dfrac{2 \sqrt{3}+3}{3}\),
当且仅当 \(t=\sqrt{3} \Rightarrow a=\sqrt{3}-1, b=2-\sqrt{3}\)时,等号成立,
即所求最大值为 \(\dfrac{2 \sqrt{3}+3}{3}\).
【C组---拓展题】
1.设\(x>0、y>0、z>0\),则三个数 \(\dfrac{1}{x}+4 y, \dfrac{1}{y}+4 z, \dfrac{1}{z}+4 x\)( )
A.都大于\(4\) \(\qquad \qquad\) B.至少有一个大于\(4\) \(\qquad \qquad\) C.至少有一个不小于\(4\) \(\qquad \qquad\) D.至少有一个不大于\(4\)
2.已知\(a>b>c\),若 \(\dfrac{1}{a-b}+\dfrac{4}{b-c} \geq \dfrac{m}{a-c}\)恒成立,则\(m\)的最大值为( )
A.\(3\) \(\qquad \qquad\) B.\(4\) \(\qquad \qquad\) C.\(8\) \(\qquad \qquad\) D.\(9\)
3.设实数\(x,y\)满足 \(\dfrac{x^{2}}{4}-y^{2}=1\),则\(3x^2-2xy\)的最小值是\(\underline{\quad \quad}\).
4.设\(a ,b ,c\)均为正数,且\(a+b+c=1\).证明:
(1) \(a b+b c+a c \leq \dfrac{1}{3}\); (2) \(\dfrac{a^{2}}{b}+\dfrac{b^{2}}{c}+\dfrac{c^{2}}{a} \geq 1\).
参考答案
-
答案 \(C\)
解析 假设三个数 \(\dfrac{1}{x}+4 y<4\)且 \(\dfrac{1}{y}+4 z<4\)且 \(\dfrac{1}{z}+4 x<4\),
相加得: \(\dfrac{1}{x}+4 x+\dfrac{1}{y}+4 y+\dfrac{1}{z}+4 z<12\),
由基本不等式得: \(\dfrac{1}{x}+4 x \geq 4 ; \quad \dfrac{1}{y}+4 y \geq 4 ; \dfrac{1}{z}+4 z \geq 4\);
相加得: \(\dfrac{1}{x}+4 x+\dfrac{1}{y}+4 y+\dfrac{1}{z}+4 z \geq 12\),与假设矛盾;
所以假设不成立,三个数 \(\dfrac{1}{x}+4 y, \dfrac{1}{y}+4 z, \dfrac{1}{z}+4 x\)至少有一个不小于\(4\).
故选:\(C\). -
答案 \(D\)
解析 由\(a>b>c\),知\(a-b>0,b-c>0,a-c>0\),
由 \(\dfrac{1}{a-b}+\dfrac{4}{b-c} \geq \dfrac{m}{a-c}\),得 \(m \leq(a-c)\left(\dfrac{1}{a-b}+\dfrac{4}{b-c}\right)\),
又\(∵a-c=a-b+b-c\),
\(\left.\therefore(a-c)\left(\dfrac{1}{a-b}+\dfrac{4}{b-c}\right)=[(a-b)+(b-c)]\left(\dfrac{1}{a-b}+\dfrac{4}{b-c}\right)\right]\)
\(=5+\dfrac{4(a-b)}{b-c}+\dfrac{b-c}{a-b} \geq 5+2 \sqrt{\dfrac{4(a-b)}{b-c} \cdot \dfrac{b-c}{a-b}}=9\),当且仅当 \(\dfrac{4(a-b)}{b-c}=\dfrac{b-c}{a-b}\),
即\(b-c=2(a-b)\)时, \(\left.(a-c)\left(\dfrac{1}{a-b}+\dfrac{4}{b-c}\right)\right]\)取得最小值\(9\),
\(∴m≤9\),\(∴m\)的最大值为\(9\).
故选:\(D\). -
答案 \(6+4 \sqrt{2}\)
解析方法1 \(3 x^{2}-2 x y=\dfrac{3 x^{2}-2 x y}{\dfrac{x^{2}}{4}-y^{2}}=\dfrac{3-\dfrac{2 y}{x}}{\dfrac{1}{4}-\left(\dfrac{y}{x}\right)^{2}}\)
令 \(t=\dfrac{y}{x}\), \(\because \dfrac{x^{2}}{4}-y^{2}=1\)
\(\therefore \dfrac{x^{2}}{4}-t^{2} x^{2}=1 \Rightarrow t^{2}=\dfrac{1}{4}-\dfrac{1}{x^{2}}<\dfrac{1}{4} \Rightarrow-\dfrac{1}{2}<t<\dfrac{1}{2}\),
则 \(3 x^{2}-2 x y=\dfrac{3-2 t}{\dfrac{1}{4}-t^{2}}\)
再令\(u=3-2t (2<u<4)\)
则 \(\dfrac{3}{x^{2}}-2 x y=\dfrac{u}{\dfrac{1}{4}-\left(\dfrac{3-u}{2}\right)^{2}}=\dfrac{4 u}{-u^{2}+6 u-8}=\dfrac{4}{-\left(u+\dfrac{8}{u}\right)+6}\)\(\geq \dfrac{4}{-4 \sqrt{2}+6}=6+4 \sqrt{2}\)
当且仅当 \(u=2 \sqrt{2}\)时取到等号,
方法2 \(\because \dfrac{x^{2}}{4}-y^{2}=1\) \(\therefore\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}+y\right)=1\)
令 \(t=\dfrac{x}{2}+y\),则 \(\dfrac{x}{2}-y=\dfrac{1}{t}\),
\(\therefore x=t+\dfrac{1}{t}, y=\dfrac{1}{2}\left(t-\dfrac{1}{t}\right)\)
\(\therefore 3 x^{2}-2 x y=3\left(t+\dfrac{1}{t}\right)^{2}-2\left(t+\dfrac{1}{t}\right)\left(t-\dfrac{1}{t}\right)\)\(=2 t^{2}+\dfrac{4}{t^{2}}+6 \geq 4 \sqrt{2}+6=6+4 \sqrt{2}\)
当且仅当 \(t^{2}=\sqrt{2}\)时取到等号. -
证明 :(1)由 \(a^2+b^2≥2ab ,b^2+c^2≥2bc ,c^2+a^2≥2ca\).
得 \(a^2+b^2+c^2≥ab+bc+ca\).
由题设得\((a+b+c)^2=1\),
即 \(a^2+b^2+c^2+2ab+2bc+2ca=1\).
所以 \(3(ab+bc+ca)≤1\),
即 \(a b+b c+c a \leq \dfrac{1}{3}\).
(2)因为 \(\dfrac{a^{2}}{b}+b \geq 2 a, \dfrac{b^{2}}{c}+c \geq 2 b, \dfrac{c^{2}}{a}+a \geq 2 c\),
故 \(\dfrac{a^{2}}{b}+\dfrac{b^{2}}{c}+\dfrac{c^{2}}{a}+(a+b+c) \geq 2(a+b+c)\),
即 \(\dfrac{a^{2}}{b}+\dfrac{b^{2}}{c}+\dfrac{c^{2}}{a} \geq a+b+c\),
所以 \(\dfrac{a^{2}}{b}+\dfrac{b^{2}}{c}+\dfrac{c^{2}}{a} \geq 1\).