希望快速了解或快速回顧高中數學的讀者可以只看基礎知識部分。其餘部分是為需要參加學科考試或需要一定知識提升的讀者準備的。
先前我們已經了解，任何實數的平方或取實數值的代數式的平方一定不會小於0，這是推導各種常用不等式的起點。
定義：對於任意的2個正數，稱
a
+
b
2
{\displaystyle {\frac {a+b}{2}}}
算術平均數（arithmetic mean） ，
a
b
{\displaystyle {\sqrt {ab}}}
幾何平均數（geometric mean） 。 [ 1]
[ 1]
a
+
b
2
≥
a
b
{\displaystyle {\frac {a+b}{2}}\geq {\sqrt {ab}}}
[ 1] 算術-幾何平均值不等式（inequality of arithmetic and geometric means） ，簡稱平均值不等式 或基本不等式 。
證明：
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}}}
算術-幾何平均值不等式的幾何意義。 從幾何角度看，基本不等式本身的意義如右圖所示。另一方面，從代數角度來看，平均值不等式可以說明對於2個正數：
如果限定它們的和為固定值，那麼它們的乘積一定存在最大取值
a
b
{\displaystyle {\sqrt {ab}}}
如果限定它們的乘積為固定值，那麼它們的和一定存在最小取值
a
+
b
2
{\displaystyle {\frac {a+b}{2}}}
常用以口訣簡記為「和定積最大，積定和最小」。「和定積最大」的意義是說當長方形的周長一定時，其面積有最大值，當且僅當長方形是正方形時面積值達到最大；「積定和最小」的意義是說當長方形的面積一定時，其周長有最小值，當且僅當長方形是正方形時周長值達到最大。
a
+
b
2
{\displaystyle {\frac {a+b}{2}}}
a
b
{\displaystyle {\sqrt {ab}}}
任意兩個正數的等差中項不小於它們的等比中項 。[ 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關鍵是拼湊出「定和」或「定積」的形式，並保證等號要能成立。常用技巧包括：
加項或拆項。
減少變量數目或統一變元
平方後再利用基本不等式
利用約束條件中的常量代換 有時候，要求最值的式子和所給的定值約束條件不太相符。這時可以考慮利用約束條件可以進行變量或常量的等價替換，以便拼湊出滿足基本不等式的形式條件。
通過換元，將包含多個變量的式子減少變量數目或統一變元。
常量代換，這種方法最常用於「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}}
證明：
{
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}}}
[ 編輯 ]
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}})}
對於2個正數a和b，有下列不等式鏈成立：
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}\}}
它可以看作平均數不等式 在只涉及2個變元時的特殊情形，給出了4種最常見的平均數的大小順序。
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}}}
上述均值不等式鏈的一部分——對數均值不等式 （ALG不等式）在解決函數的極值點偏移問題 上有着重要的應用。
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，請用平均值不等式說明當x取何值時，
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}}}}}
判斷下列做法是否有誤。如果有誤，請指出具體出錯原因。 (1)設x > 0，因為
(
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)設k > 1，因為
(
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)設x > 0，因為
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)因為
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：任意兩個正數的算術平均數不小於它們的幾何平均數[1]。即對於任意的正數a和b，一定有：
注意：(1)只有保證a和b都是正數的前提下，才能套用平均值不等式。(2)不等式的取等條件是否成立非常重要，在證明題和解答題中都需要單獨說明。
提示：對於了解過等差數列和等比數列概念的讀者，如果把看作是正數a和b的等差中項，看作是它們的等比中項，那麼這個基本不等式也可以表述為任意兩個正數的等差中項不小於它們的等比中項。[1]
相關例題1：
已知x > 0，求函數的最小值。
視野拓展：不等式鏈可以擴充到令人大跌眼鏡的程度。例如對於，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)
