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
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n
b
=
k
{\displaystyle ma+nb=k}
1
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1
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}}
m
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n
b
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1
{\displaystyle {\frac {m}{a}}+{\frac {n}{b}}=1}
a
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{\displaystyle a+b}
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b
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2
{\displaystyle a+b=2}
1
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1
b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}}
1
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9
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1
{\displaystyle {\frac {1}{a}}+{\frac {9}{b}}=1}
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b
{\displaystyle a+b}
a
2
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3
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b
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1
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0
{\displaystyle a^{2}+3ab-1=0}
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b
{\displaystyle a+b}
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3
b
−
2
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0
{\displaystyle a+3b-2=0}
3
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27
b
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1
{\displaystyle 3^{a}+27^{b}+1}
a
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2
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2
a
b
=
8
{\displaystyle a+2b+2ab=8}
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2
b
{\displaystyle a+2b}
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2
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3
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4
b
2
−
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0
{\displaystyle a^{2}-3ab+4b^{2}-c=0}
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b
{\displaystyle {\frac {c}{ab}}}
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2
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−
c
{\displaystyle a+2b-c}
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3
=
2
{\displaystyle {\frac {a}{2}}+{\frac {b}{3}}=2}
a
b
{\displaystyle ab}
a
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b
=
4
{\displaystyle a+b=4}
1
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1
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1
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3
{\displaystyle {\frac {1}{a+1}}+{\frac {1}{b+3}}}
1
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1
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2
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b
{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}+2{\sqrt {ab}}}
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{\displaystyle x^{2}(2-x)}
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≥
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{\displaystyle (a+b)(b+c)(a+c)\geq 8abc}
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{\displaystyle a,b,c>0}
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{\displaystyle a^{2}+b^{2}+c^{2}\geq ab+bc+ca}
3
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≥
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2
{\displaystyle 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}}
证明：
{
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{\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
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{\displaystyle a=b=c}
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{\displaystyle {\frac {a^{2}}{b}}+{\frac {b^{2}}{c}}+{\frac {c^{2}}{a}}\geq a+b+c}
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1
{\displaystyle a+b+c=1}
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1
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≥
8
{\displaystyle ({\frac {1}{a}}-1)({\frac {1}{b}}-1)({\frac {1}{c}}-1)\geq 8}
a
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=
1
{\displaystyle abc=1}
a
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1
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{\displaystyle {\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}\geq {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}
编辑
y
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{\displaystyle y=e^{x}+e^{-x}}
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{\displaystyle y={\frac {\sin x}{2}}+{\frac {2}{\sin x}}\quad (0<x<\pi )}
lg
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lg
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{\displaystyle \lg 2^{a}+\lg 8^{b}=\lg 2}
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3
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{\displaystyle {\frac {1}{a}}+{\frac {1}{3b}}}
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{\displaystyle 2^{a}+2^{b}=1}
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{\displaystyle a+b}
lg
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×
lg
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{\displaystyle \lg 9\times \lg 11}
1
{\displaystyle 1}
2
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5
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=
20
{\displaystyle 2a+5b=20}
lg
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lg
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{\displaystyle \lg a+\lg b}
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{\displaystyle {\frac {1}{a}}+{\frac {1}{b}}}
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lg
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{\displaystyle f(x)=\lg x\quad (x\in \mathbb {R} ^{+})}
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{\displaystyle x_{1},x_{2}\in (-0,+\infty )}
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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}})}
对于2个正数a和b，有下列不等式链成立：
min
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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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{\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
+
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{\displaystyle {\frac {a+b}{2}}}
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{\displaystyle {\sqrt {ab}}}
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{\displaystyle {\sqrt {\frac {a^{2}+b^{2}}{2}}}}
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{\displaystyle {\frac {2ab}{a+b}}}
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{\displaystyle 0<a\leq b}
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ln
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ln
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<
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<
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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。
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)
