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
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+
1
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}}
m
a
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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
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2
−
x
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{\displaystyle x^{2}(2-x)}
(
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b
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(
b
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c
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≥
8
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b
c
{\displaystyle (a+b)(b+c)(a+c)\geq 8abc}
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,
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c
>
0
{\displaystyle a,b,c>0}
a
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b
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2
≥
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+
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c
a
{\displaystyle a^{2}+b^{2}+c^{2}\geq ab+bc+ca}
3
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2
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≥
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2
{\displaystyle 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}}
證明：
{
a
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b
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⇒
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+
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+
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≥
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+
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⇒
3
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≥
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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
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2
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≥
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{\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
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{\displaystyle {\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}\geq {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}
編輯
y
=
e
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e
−
x
{\displaystyle y=e^{x}+e^{-x}}
y
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sin
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sin
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{\displaystyle y={\frac {\sin x}{2}}+{\frac {2}{\sin x}}\quad (0<x<\pi )}
lg
2
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lg
8
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=
lg
2
{\displaystyle \lg 2^{a}+\lg 8^{b}=\lg 2}
1
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+
1
3
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{\displaystyle {\frac {1}{a}}+{\frac {1}{3b}}}
2
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+
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
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+
1
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{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}}
f
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lg
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{\displaystyle f(x)=\lg x\quad (x\in \mathbb {R} ^{+})}
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∈
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{\displaystyle x_{1},x_{2}\in (-0,+\infty )}
f
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{\displaystyle {\frac {f(x_{1})+f(x_{2})}{2}}}
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{\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
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max
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{\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
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≤
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≤
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max
{
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b
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}
{\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
<
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≤
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{\displaystyle 0<a\leq b}
a
<
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)
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<
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<
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<
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<
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2
<
b
−
a
ln
b
−
ln
a
<
(
a
3
+
b
3
2
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3
<
(
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+
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2
)
2
<
a
+
a
b
+
b
3
<
(
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2
3
+
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3
2
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3
2
<
1
e
(
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a
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)
1
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−
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<
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+
b
2
<
2
3
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+
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2
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+
b
<
a
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+
b
2
2
<
(
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b
b
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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 上有著重要的應用。
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)
