希望快速了解或快速回顾高中数学的读者可以只看基础知识部分。其余部分是为需要参加学科考试或需要一定知识提升的读者准备的。
先前我们已经了解，任何实数的平方或取实数值的代数式的平方一定不会小于0，这是推导各种常用不等式的起点。
定义：对于任意的2个正数，称
a
+
b
2
{\displaystyle {\frac {a+b}{2}}}
算术平均数（arithmetic mean） ，
a
b
{\displaystyle {\sqrt {ab}}}
几何平均数（geometric mean） 。 [ 1]
[ 1]
a
+
b
2
≥
a
b
{\displaystyle {\frac {a+b}{2}}\geq {\sqrt {ab}}}
[ 1] 算术-几何平均值不等式（inequality of arithmetic and geometric means） ，简称平均值不等式 或基本不等式 。
证明：
a
+
b
2
≥
a
b
>
0
⇔
a
+
b
≥
2
a
b
>
0
⇔
(
a
+
b
)
2
≥
4
a
b
>
0
⇔
a
2
+
2
a
b
+
b
2
≥
4
a
b
>
0
⇔
a
2
+
2
a
b
+
b
2
−
4
a
b
≥
0
⇔
a
2
−
2
a
b
+
b
2
≥
0
⇔
(
a
−
b
)
2
≥
0
{\displaystyle {\begin{aligned}{\frac {a+b}{2}}\geq {\sqrt {ab}}>0&\Leftrightarrow a+b\geq 2{\sqrt {ab}}>0\\&\Leftrightarrow (a+b)^{2}\geq 4ab>0\\&\Leftrightarrow a^{2}+2ab+b^{2}\geq 4ab>0\\&\Leftrightarrow a^{2}+2ab+b^{2}-4ab\geq 0\\&\Leftrightarrow a^{2}-2ab+b^{2}\geq 0\\&\Leftrightarrow (a-b)^{2}\geq 0\\\end{aligned}}}
算术-几何平均值不等式的几何意义。 从几何角度看，基本不等式本身的意义如右图所示。另一方面，从代数角度来看，平均值不等式可以说明对于2个正数：
如果限定它们的和为固定值，那么它们的乘积一定存在最大取值
a
b
{\displaystyle {\sqrt {ab}}}
如果限定它们的乘积为固定值，那么它们的和一定存在最小取值
a
+
b
2
{\displaystyle {\frac {a+b}{2}}}
常用以口诀简记为“和定积最大，积定和最小”。“和定积最大”的意义是说当长方形的周长一定时，其面积有最大值，当且仅当长方形是正方形时面积值达到最大；“积定和最小”的意义是说当长方形的面积一定时，其周长有最小值，当且仅当长方形是正方形时周长值达到最大。
a
+
b
2
{\displaystyle {\frac {a+b}{2}}}
a
b
{\displaystyle {\sqrt {ab}}}
任意两个正数的等差中项不小于它们的等比中项 。[ 1]
y
=
x
+
1
x
{\displaystyle y=x+{\frac {1}{x}}}
答案：
2
{\displaystyle 2}
y
=
x
+
x
2
+
1
x
{\displaystyle y=x+{\frac {x^{2}+1}{x}}}
y
=
x
+
x
+
3
x
−
1
{\displaystyle y=x+{\frac {x+3}{x-1}}}
y
=
x
2
+
1
x
−
1
{\displaystyle y={\frac {x^{2}+1}{x-1}}}
y
=
x
x
2
+
4
{\displaystyle y={\frac {x}{x^{2}+4}}}
(
a
−
b
)
(
b
−
c
)
{\displaystyle {\sqrt {(a-b)(b-c)}}}
a
−
c
2
{\displaystyle {\frac {a-c}{2}}}
y
=
x
+
4
x
−
3
{\displaystyle y=x+{\frac {4}{x-3}}}
x
>
1
2
{\displaystyle x>{\frac {1}{2}}}
y
=
2
x
2
+
x
+
1
2
x
−
1
{\displaystyle y={\frac {2x^{2}+x+1}{2x-1}}}
y
=
3
−
x
x
2
+
x
+
1
{\displaystyle y={\frac {3-x}{x^{2}+x+1}}}
y
=
3
x
2
+
6
x
2
+
1
{\displaystyle y=3x^{2}+{\frac {6}{x^{2}+1}}}
y
=
x
(
3
−
x
)
{\displaystyle y=x(3-x)}
y
1
=
(
x
−
1
)
(
x
+
2
)
(
−
2
≤
x
≤
1
)
{\displaystyle y_{1}=(x-1)(x+2)\quad (-2\leq x\leq 1)}
y
2
=
(
x
−
1
)
(
x
−
2
)
(
1
≤
x
≤
2
)
{\displaystyle y_{2}=(x-1)(x-2)\quad (1\leq x\leq 2)}
y
3
=
(
x
+
1
)
(
x
+
2
)
(
−
2
≤
x
≤
−
1
)
{\displaystyle y_{3}=(x+1)(x+2)\quad (-2\leq x\leq -1)}
y
4
=
(
x
−
1
)
(
2
x
+
1
)
(
−
2
≤
x
≤
−
1
)
{\displaystyle y_{4}=(x-1)(2x+1)\quad (-2\leq x\leq -1)}
y
=
1
x
−
1
x
−
3
{\displaystyle y={\frac {1}{x}}-{\frac {1}{x-3}}}
使用基本不等式求最值的关键是拼凑出“定和”或“定积”的形式，并保证等号要能成立。常用技巧包括：
加项或拆项。
减少变量数目或统一变元
平方后再利用基本不等式
利用约束条件中的常量代换 有时候，要求最值的式子和所给的定值约束条件不太相符。这时可以考虑利用约束条件可以进行变量或常量的等价替换，以便拼凑出满足基本不等式的形式条件。
通过换元，将包含多个变量的式子减少变量数目或统一变元。
常量代换，这种方法最常用于“已知
m
a
+
n
b
=
k
{\displaystyle ma+nb=k}
1
a
+
1
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}}
m
a
+
n
b
=
1
{\displaystyle {\frac {m}{a}}+{\frac {n}{b}}=1}
a
+
b
{\displaystyle a+b}
a
+
b
=
2
{\displaystyle a+b=2}
1
a
+
1
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}}
1
a
+
9
b
=
1
{\displaystyle {\frac {1}{a}}+{\frac {9}{b}}=1}
a
+
b
{\displaystyle a+b}
a
2
+
3
a
b
−
1
=
0
{\displaystyle a^{2}+3ab-1=0}
a
+
b
{\displaystyle a+b}
a
+
3
b
−
2
=
0
{\displaystyle a+3b-2=0}
3
a
+
27
b
+
1
{\displaystyle 3^{a}+27^{b}+1}
a
+
2
b
+
2
a
b
=
8
{\displaystyle a+2b+2ab=8}
a
+
2
b
{\displaystyle a+2b}
a
2
−
3
a
b
+
4
b
2
−
c
=
0
{\displaystyle a^{2}-3ab+4b^{2}-c=0}
c
a
b
{\displaystyle {\frac {c}{ab}}}
a
+
2
b
−
c
{\displaystyle a+2b-c}
a
2
+
b
3
=
2
{\displaystyle {\frac {a}{2}}+{\frac {b}{3}}=2}
a
b
{\displaystyle ab}
a
+
b
=
4
{\displaystyle a+b=4}
1
a
+
1
+
1
b
+
3
{\displaystyle {\frac {1}{a+1}}+{\frac {1}{b+3}}}
1
a
+
1
b
+
2
a
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}+2{\sqrt {ab}}}
x
2
(
2
−
x
)
{\displaystyle x^{2}(2-x)}
(
a
+
b
)
(
b
+
c
)
(
a
+
c
)
≥
8
a
b
c
{\displaystyle (a+b)(b+c)(a+c)\geq 8abc}
a
,
b
,
c
>
0
{\displaystyle a,b,c>0}
a
2
+
b
2
+
c
2
≥
a
b
+
b
c
+
c
a
{\displaystyle a^{2}+b^{2}+c^{2}\geq ab+bc+ca}
3
(
a
2
+
b
2
+
c
2
)
≥
(
a
+
b
+
c
)
2
{\displaystyle 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}}
证明：
{
a
2
+
b
2
≥
2
a
b
b
2
+
c
2
≥
2
b
c
c
2
+
a
2
≥
2
c
a
⇒
(
a
2
+
b
2
)
+
(
b
2
+
c
2
)
+
(
c
2
+
a
2
)
≥
2
a
b
+
2
b
c
+
2
c
a
⇒
2
(
a
2
+
b
2
+
c
2
)
≥
2
(
a
b
+
b
c
+
c
a
)
⇒
2
(
a
2
+
b
2
+
c
2
)
+
(
a
2
+
b
2
+
c
2
)
≥
2
(
a
b
+
b
c
+
c
a
)
+
(
a
2
+
b
2
+
c
2
)
⇒
3
(
a
2
+
b
2
+
c
2
)
≥
(
a
+
b
+
c
)
2
{\displaystyle {\begin{array}{l}\left\{{\begin{array}{l}a^{2}+b^{2}\geq 2ab\\b^{2}+c^{2}\geq 2bc\\c^{2}+a^{2}\geq 2ca\end{array}}\right.\\\Rightarrow (a^{2}+b^{2})+(b^{2}+c^{2})+(c^{2}+a^{2})\geq 2ab+2bc+2ca\\\Rightarrow 2(a^{2}+b^{2}+c^{2})\geq 2(ab+bc+ca)\\\Rightarrow 2(a^{2}+b^{2}+c^{2})+(a^{2}+b^{2}+c^{2})\geq 2(ab+bc+ca)+(a^{2}+b^{2}+c^{2})\\\Rightarrow 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}\end{array}}}
a
=
b
=
c
{\displaystyle a=b=c}
a
2
b
+
b
2
c
+
c
2
a
≥
a
+
b
+
c
{\displaystyle {\frac {a^{2}}{b}}+{\frac {b^{2}}{c}}+{\frac {c^{2}}{a}}\geq a+b+c}
a
+
b
+
c
=
1
{\displaystyle a+b+c=1}
(
1
a
−
1
)
(
1
b
−
1
)
(
1
c
−
1
)
≥
8
{\displaystyle ({\frac {1}{a}}-1)({\frac {1}{b}}-1)({\frac {1}{c}}-1)\geq 8}
a
b
c
=
1
{\displaystyle abc=1}
a
+
b
+
c
≥
1
a
+
1
b
+
1
c
{\displaystyle {\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}\geq {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}
[ 编辑 ]
y
=
e
x
+
e
−
x
{\displaystyle y=e^{x}+e^{-x}}
y
=
sin
x
2
+
2
sin
x
(
0
<
x
<
π
)
{\displaystyle y={\frac {\sin x}{2}}+{\frac {2}{\sin x}}\quad (0<x<\pi )}
lg
2
a
+
lg
8
b
=
lg
2
{\displaystyle \lg 2^{a}+\lg 8^{b}=\lg 2}
1
a
+
1
3
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{3b}}}
2
a
+
2
b
=
1
{\displaystyle 2^{a}+2^{b}=1}
a
+
b
{\displaystyle a+b}
lg
9
×
lg
11
{\displaystyle \lg 9\times \lg 11}
1
{\displaystyle 1}
2
a
+
5
b
=
20
{\displaystyle 2a+5b=20}
lg
a
+
lg
b
{\displaystyle \lg a+\lg b}
1
a
+
1
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}}
f
(
x
)
=
lg
x
(
x
∈
R
+
)
{\displaystyle f(x)=\lg x\quad (x\in \mathbb {R} ^{+})}
x
1
,
x
2
∈
(
−
0
,
+
∞
)
{\displaystyle x_{1},x_{2}\in (-0,+\infty )}
f
(
x
1
)
+
f
(
x
2
)
2
{\displaystyle {\frac {f(x_{1})+f(x_{2})}{2}}}
f
(
x
1
+
x
2
2
)
{\displaystyle f({\frac {x_{1}+x_{2}}{2}})}
对于2个正数a和b，有下列不等式链成立：
min
{
a
,
b
}
≤
2
1
a
+
1
b
≤
a
b
≤
a
+
b
2
≤
a
2
+
b
2
2
≤
max
{
a
,
b
}
{\displaystyle \min\{a,b\}\leq {\frac {2}{{\frac {1}{a}}+{\frac {1}{b}}}}\leq {\sqrt {ab}}\leq {\frac {a+b}{2}}\leq {\sqrt {\frac {a^{2}+b^{2}}{2}}}\leq \max\{a,b\}}
min
{
a
2
,
b
2
}
≤
(
2
1
a
+
1
b
)
2
≤
a
b
≤
(
a
+
b
2
)
2
≤
a
2
+
b
2
2
≤
max
{
a
2
,
b
2
}
{\displaystyle \min\{a^{2},b^{2}\}\leq ({\frac {2}{{\frac {1}{a}}+{\frac {1}{b}}}})^{2}\leq ab\leq ({\frac {a+b}{2}})^{2}\leq {\frac {a^{2}+b^{2}}{2}}\leq \max\{a^{2},b^{2}\}}
它可以看作平均数不等式 在只涉及2个变元时的特殊情形，给出了4种最常见的平均数的大小顺序。
a
+
b
2
{\displaystyle {\frac {a+b}{2}}}
a
b
{\displaystyle {\sqrt {ab}}}
a
2
+
b
2
2
{\displaystyle {\sqrt {\frac {a^{2}+b^{2}}{2}}}}
2
a
b
a
+
b
{\displaystyle {\frac {2ab}{a+b}}}
0
<
a
≤
b
{\displaystyle 0<a\leq b}
a
<
(
a
b
b
a
)
1
a
+
b
<
2
a
2
b
2
a
2
+
b
2
<
2
a
b
a
+
b
<
a
b
<
(
a
b
)
1
−
1
3
(
a
1
3
+
b
1
3
)
2
<
b
−
a
ln
b
−
ln
a
<
(
a
3
+
b
3
2
)
3
<
(
a
+
b
2
)
2
<
a
+
a
b
+
b
3
<
(
a
2
3
+
b
2
3
2
)
3
2
<
1
e
(
a
a
b
b
)
1
a
−
b
<
a
+
b
2
<
2
3
a
2
+
a
b
+
b
2
a
+
b
<
a
2
+
b
2
2
<
(
a
a
b
b
)
1
a
+
b
<
a
2
+
b
2
a
+
b
<
a
2
−
a
b
+
b
2
<
b
{\displaystyle {\begin{aligned}a<(a^{b}b^{a})^{\frac {1}{a+b}}<{\sqrt {\frac {2a^{2}b^{2}}{a^{2}+b^{2}}}}<{\frac {2ab}{a+b}}<{\sqrt {ab}}<{\frac {(ab)^{1-{\frac {1}{\sqrt {3}}}}(a^{\frac {1}{\sqrt {3}}}+b^{\frac {1}{\sqrt {3}}})}{2}}<{\frac {b-a}{\ln b-\ln a}}\\<({\frac {{\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}}{2}})^{3}<({\frac {{\sqrt {a}}+{\sqrt {b}}}{2}})^{2}<{\frac {a+{\sqrt {ab}}+b}{3}}<({\frac {{\sqrt[{3}]{a^{2}}}+{\sqrt[{3}]{b^{2}}}}{2}})^{\frac {3}{2}}<{\frac {1}{e}}({\frac {a^{a}}{b^{b}}})^{\frac {1}{a-b}}\\<{\frac {a+b}{2}}<{\frac {2}{3}}{\frac {a^{2}+ab+b^{2}}{a+b}}<{\sqrt {\frac {a^{2}+b^{2}}{2}}}<(a^{a}b^{b})^{\frac {1}{a+b}}<{\frac {a^{2}+b^{2}}{a+b}}<{\sqrt {a^{2}-ab+b^{2}}}<b\end{aligned}}}
上述均值不等式链的一部分——对数均值不等式 （ALG不等式）在解决函数的极值点偏移问题 上有着重要的应用。
已知
a
,
b
>
0
{\displaystyle a,b>0}
a
b
≥
4
(
a
−
a
b
)
{\displaystyle {\frac {a}{b}}\geq 4(a-ab)}
已知
a
,
b
>
0
,
a
+
2
b
=
1
{\displaystyle a,b>0,a+2b=1}
a
b
≤
1
8
{\displaystyle ab\leq {\frac {1}{8}}}
已知0 < x < 1，请用平均值不等式说明当x取何值时，
x
(
1
−
x
)
{\displaystyle {\sqrt {x(1-x)}}}
已知0 < x < 2，求
x
(
2
−
x
)
2
{\displaystyle {\sqrt {x(2-x)^{2}}}}
已知x > 0，求
(
x
+
1
x
)
−
1
x
+
1
x
{\displaystyle (x+{\frac {1}{x}})-{\frac {1}{x+{\frac {1}{x}}}}}
判断下列做法是否有误。如果有误，请指出具体出错原因。 (1)设x > 0，因为
(
x
+
1
x
)
+
1
x
+
1
x
≥
2
(
x
+
1
x
)
⋅
1
x
+
1
x
=
2
1
=
2
{\displaystyle (x+{\frac {1}{x}})+{\frac {1}{x+{\frac {1}{x}}}}\geq 2{\sqrt {(x+{\frac {1}{x}})\cdot {\frac {1}{x+{\frac {1}{x}}}}}}=2{\sqrt {1}}=2}
(
x
+
1
x
)
+
1
x
+
1
x
{\displaystyle (x+{\frac {1}{x}})+{\frac {1}{x+{\frac {1}{x}}}}}
(2)设k > 1，因为
(
1
−
1
k
)
(
1
+
2
k
)
=
1
2
(
1
−
1
k
)
(
2
+
1
k
)
≤
1
2
(
(
1
−
1
k
)
+
(
2
+
1
k
)
2
)
2
=
1
2
(
3
2
)
2
=
9
8
{\displaystyle (1-{\frac {1}{k}})(1+{\frac {2}{k}})={\frac {1}{2}}(1-{\frac {1}{k}})(2+{\frac {1}{k}})\leq {\frac {1}{2}}({\frac {(1-{\frac {1}{k}})+(2+{\frac {1}{k}})}{2}})^{2}={\frac {1}{2}}({\frac {3}{2}})^{2}={\frac {9}{8}}}
(
1
−
1
k
)
(
1
+
2
k
)
{\displaystyle (1-{\frac {1}{k}})(1+{\frac {2}{k}})}
9
8
{\displaystyle {\frac {9}{8}}}
(3)设x > 0，因为
2
x
(
x
+
1
)
≤
(
(
2
x
)
+
(
x
+
1
)
2
)
2
=
(
3
x
+
1
2
)
2
{\displaystyle 2x(x+1)\leq ({\frac {(2x)+(x+1)}{2}})^{2}=({\frac {3x+1}{2}})^{2}}
(
3
x
+
1
)
2
2
x
(
x
+
1
)
≥
(
3
x
+
1
)
2
(
3
x
+
1
2
)
2
=
4
{\displaystyle {\frac {(3x+1)^{2}}{2x(x+1)}}\geq {\frac {(3x+1)^{2}}{({\frac {3x+1}{2}})^{2}}}=4}
(
3
x
+
1
)
2
2
x
(
x
+
1
)
{\displaystyle {\frac {(3x+1)^{2}}{2x(x+1)}}}
(4)因为
3
2
x
(
1
−
x
)
(
2
−
x
)
3
≤
2
x
+
(
1
−
x
)
+
(
2
−
x
)
=
3
{\displaystyle 3{\sqrt[{3}]{2x(1-x)(2-x)}}\leq 2x+(1-x)+(2-x)=3}
x
(
1
−
x
)
(
2
−
x
)
{\displaystyle x(1-x)(2-x)}
定理：任意两个正数的算术平均数不小于它们的几何平均数[1]。即对于任意的正数a和b，一定有：
注意：(1)只有保证a和b都是正数的前提下，才能套用平均值不等式。(2)不等式的取等条件是否成立非常重要，在证明题和解答题中都需要单独说明。
提示：对于了解过等差数列和等比数列概念的读者，如果把看作是正数a和b的等差中项，看作是它们的等比中项，那么这个基本不等式也可以表述为任意两个正数的等差中项不小于它们的等比中项。[1]
相关例题1：
已知x > 0，求函数的最小值。
视野拓展：不等式链可以扩充到令人大跌眼镜的程度。例如对于，可以得到：
![{\displaystyle {\begin{aligned}a<(a^{b}b^{a})^{\frac {1}{a+b}}<{\sqrt {\frac {2a^{2}b^{2}}{a^{2}+b^{2}}}}<{\frac {2ab}{a+b}}<{\sqrt {ab}}<{\frac {(ab)^{1-{\frac {1}{\sqrt {3}}}}(a^{\frac {1}{\sqrt {3}}}+b^{\frac {1}{\sqrt {3}}})}{2}}<{\frac {b-a}{\ln b-\ln a}}\\<({\frac {{\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}}{2}})^{3}<({\frac {{\sqrt {a}}+{\sqrt {b}}}{2}})^{2}<{\frac {a+{\sqrt {ab}}+b}{3}}<({\frac {{\sqrt[{3}]{a^{2}}}+{\sqrt[{3}]{b^{2}}}}{2}})^{\frac {3}{2}}<{\frac {1}{e}}({\frac {a^{a}}{b^{b}}})^{\frac {1}{a-b}}\\<{\frac {a+b}{2}}<{\frac {2}{3}}{\frac {a^{2}+ab+b^{2}}{a+b}}<{\sqrt {\frac {a^{2}+b^{2}}{2}}}<(a^{a}b^{b})^{\frac {1}{a+b}}<{\frac {a^{2}+b^{2}}{a+b}}<{\sqrt {a^{2}-ab+b^{2}}}<b\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e41df4d00773ceb0c8a09ac9899f1e18492cea84)
![{\displaystyle 3{\sqrt[{3}]{2x(1-x)(2-x)}}\leq 2x+(1-x)+(2-x)=3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f4af33b3d4fde297b661e3002810e2cc2da3297)
