平均数不等式，或称平均值不等式、均值不等式，是数学上的一组不等式，也是基本不等式的推广。它是说：如果 x 1 , x 2 , … , x n {displaystyle x_{1},x_{2},ldots ,x_{n}} 是正数，则H n ≤ G n ≤ A n ≤ Q n {displaystyle H_{n}leq G_{n}leq A_{n}leq Q_{n}}其中： H n = n ∑ i = 1 n 1 x i = n 1 x 1 + 1 x 2 + ⋯ + 1 x n {displaystyle H_{n}={dfrac {n}{displaystyle sum _{i=1}^{n}{dfrac {1}{x_{i}}}}}={dfrac {n}{{dfrac {1}{x_{1}}}+{dfrac {1}{x_{2}}}+cdots +{dfrac {1}{x_{n}}}}}}G n = ∏ i = 1 n x i n = x 1 x 2 ⋯ x n n {displaystyle G_{n}={sqrt{prod _{i=1}^{n}x_{i}}}={sqrt{x_{1}x_{2}cdots x_{n}}}}A n = ∑ i = 1 n x i n = x 1 + x 2 + ⋯ + x n n {displaystyle A_{n}={dfrac {displaystyle sum _{i=1}^{n}x_{i}}{n}}={dfrac {x_{1}+x_{2}+cdots +x_{n}}{n}}}Q n = ∑ i = 1 n x i 2 n = x 1 2 + x 2 2 + ⋯ + x n 2 n {displaystyle Q_{n}={sqrt {dfrac {displaystyle sum _{i=1}^{n}x_{i}^{2}}{n}}}={sqrt {dfrac {x_{1}^{2}+x_{2}^{2}+cdots +x_{n}^{2}}{n}}}}当且仅当 x 1 = x 2 = ⋯ = x n {displaystyle x_{1}=x_{2}=cdots =x_{n}} ，等号成立。即对这些正数：调和平均数 ≤ 几何平均数 ≤ 算术平均数 ≤ 平方平均数（方均根）简记为：“调几算方”关于均值不等式的证明方法有很多，数学归纳法（第一数学归纳法或反向归纳法）、拉格朗日乘数法、琴生不等式法、排序不等式法、柯西不等式法等等，都可以证明均值不等式，在这里简要介绍数学归纳法证明n维形式的均值不等式的方法：用数学归纳法证明，需要一个辅助结论。引理：设A≥0，B≥0，则 ( A + B ) n ≥ A n + n A n − 1 B {displaystyle left(A+Bright)^{n}geq A^{n}+nA^{n-1}B} ，且仅当B=0时取等号。引理的正确性较明显，条件A≥0，B≥0可以弱化为A≥0，A+B≥0，可以用数学归纳法证明。原题等价于： ( a 1 + a 2 + ⋯ + a n n ) n ≥ a 1 a 2 ⋯ a n {displaystyle left({frac {a_{1}+a_{2}+cdots +a_{n}}{n}}right)^{n}geq a_{1}a_{2}cdots a_{n}} , 当且仅当 a 1 = a 2 = ⋯ = a n {displaystyle a_{1}=a_{2}=cdots =a_{n}} 时取等号。当n=2时易证；假设当n=k时命题成立，即 ( a 1 + a 2 + ⋯ + a k k ) k ≥ a 1 a 2 ⋯ a k {displaystyle left({frac {a_{1}+a_{2}+cdots +a_{k}}{k}}right)^{k}geq a_{1}a_{2}cdots a_{k}} , 当且仅当 a 1 = a 2 = ⋯ = a k {displaystyle a_{1}=a_{2}=cdots =a_{k}} 时取等号。那么当n=k+1时，不妨设 a k + 1 {displaystyle a_{k+1}} 是 a 1 {displaystyle a_{1}} 、 a 2 ⋯ a k + 1 {displaystyle a_{2}cdots a_{k+1}} 中最大者，则 k a k + 1 ≥ a 1 + a 2 + ⋯ + a k {displaystyle ka_{k+1}geq a_{1}+a_{2}+cdots +a_{k}}设 S = a 1 + a 2 + ⋯ + a k {displaystyle S=a_{1}+a_{2}+cdots +a_{k}} ， ( a 1 + a 2 + ⋯ + a k + 1 k + 1 ) k + 1 = [ S k + k a k + 1 − S k ( k + 1 ) ] k + 1 {displaystyle left({frac {a_{1}+a_{2}+cdots +a_{k+1}}{k+1}}right)^{k+1}=left^{k+1}} ，根据引理[ S k + k a k + 1 − S k ( k + 1 ) ] k + 1 ≥ ( S k ) k + 1 + ( k + 1 ) ( S k ) k k a k + 1 − S k ( k + 1 ) = ( S k ) k a k + 1 ≥ a 1 a 2 ⋯ a k a k + 1 {displaystyle left^{k+1}geq left({frac {S}{k}}right)^{k+1}+(k+1)left({frac {S}{k}}right)^{k}{frac {ka_{k+1}-S}{k(k+1)}}=left({frac {S}{k}}right)^{k}a_{k+1}geq a_{1}a_{2}cdots a_{k}a_{k+1}} ,当且仅当 k a k + 1 − S = 0 {displaystyle ka_{k+1}-S=0} 且 a 1 = a 2 = ⋯ = a k {displaystyle a_{1}=a_{2}=cdots =a_{k}} 时，即 a 1 = a 2 = ⋯ = a k = a k + 1 {displaystyle a_{1}=a_{2}=cdots =a_{k}=a_{k+1}} 时取等号。此外，人教版高中数学教科书《选修4-5 不等式选讲》也介绍了一个运用数学归纳法的证明方法。先运用数学归纳法证明一个引理：若 n {displaystyle n} （ n {displaystyle n} 是正整数）个正数 a 1 , a 2 , . . . , a n {displaystyle a_{1},a_{2},...,a_{n}} 的乘积 a 1 a 2 . . . a n = 1 {displaystyle a_{1}a_{2}...a_{n}=1} ，则它们的和 a 1 + a 2 + . . . + a n ⩾ n {displaystyle a_{1}+a_{2}+...+a_{n}geqslant n} ，当且仅当 a 1 = a 2 = . . . = a n = 1 {displaystyle a_{1}=a_{2}=...=a_{n}=1} 时等号成立。此引理证明如下：当 n = 1 {displaystyle n=1} 时命题为：若 a = 1 {displaystyle a=1} ，则 a ⩾ 1 {displaystyle ageqslant 1} ，当且仅当 a = 1 {displaystyle a=1} 时等号成立。命题显然成立。假设当 n = k {displaystyle n=k} 时命题成立，则现在证明当 n = k + 1 {displaystyle n=k+1} 时命题也成立。若这 k + 1 {displaystyle k+1} 个数全部是1，即 a 1 = a 2 = . . . = a k + 1 = 1 {displaystyle a_{1}=a_{2}=...=a_{k+1}=1} ，则命题显然成立。若这 k + 1 {displaystyle k+1} 个数不全是1，则易证明必存在 i ≠ j {displaystyle ineq j} 使 a i > 1 , a j < 1 {displaystyle a_{i}>1,a_{j}<1} 。不妨设 a 1 > 1 , a 2 < 1 {displaystyle a_{1}>1,a_{2}<1} 。由归纳假设，因为 ( a 1 a 2 ) a 3 . . . a k + 1 = 1 {displaystyle (a_{1}a_{2})a_{3}...a_{k+1}=1} ，所以 a 1 a 2 + a 3 + . . . + a k + 1 ⩾ k {displaystyle a_{1}a_{2}+a_{3}+...+a_{k+1}geqslant k} ，记此式为①式。由 a 1 > 1 , a 2 < 1 {displaystyle a_{1}>1,a_{2}<1} ，知 a 1 − 1 > 0 , 1 − a 2 > 0 {displaystyle a_{1}-1>0,1-a_{2}>0} ，则 ( a 1 − 1 ) ( 1 − a 2 ) = a 1 + a 2 − a 1 a 2 − 1 > 0 {displaystyle (a_{1}-1)(1-a_{2})=a_{1}+a_{2}-a_{1}a_{2}-1>0} ，整理得 a 1 + a 2 > 1 + a 1 a 2 {displaystyle a_{1}+a_{2}>1+a_{1}a_{2}} ，记此式为②式。①+②得 a 1 + a 2 + a 1 a 2 + a 3 + . . . + a k + 1 > k + 1 + a 1 a 2 {displaystyle a_{1}+a_{2}+a_{1}a_{2}+a_{3}+...+a_{k+1}>k+1+a_{1}a_{2}} ，整理得 a 1 + a 2 + . . . + a k + 1 > k + 1 {displaystyle a_{1}+a_{2}+...+a_{k+1}>k+1} （此时等号不成立），命题成立。综上，由数学归纳法，引理成立。现在为了证明平均值不等式，考虑 n {displaystyle n} 个正数 a 1 a 1 a 2 . . . a n n , a 2 a 1 a 2 . . . a n n , . . . , a n a 1 a 2 . . . a n n {displaystyle {frac {a_{1}}{sqrt{a_{1}a_{2}...a_{n}}}},{frac {a_{2}}{sqrt{a_{1}a_{2}...a_{n}}}},...,{frac {a_{n}}{sqrt{a_{1}a_{2}...a_{n}}}}} ，它们的积为1，由引理，它们的和 a 1 a 1 a 2 . . . a n n + a 2 a 1 a 2 . . . a n n + . . . + a n a 1 a 2 . . . a n n = a 1 + a 2 + . . . + a n a 1 a 2 . . . a n n ⩾ n {displaystyle {frac {a_{1}}{sqrt{a_{1}a_{2}...a_{n}}}}+{frac {a_{2}}{sqrt{a_{1}a_{2}...a_{n}}}}+...+{frac {a_{n}}{sqrt{a_{1}a_{2}...a_{n}}}}={frac {a_{1}+a_{2}+...+a_{n}}{sqrt{a_{1}a_{2}...a_{n}}}}geqslant n} ，当且仅当 a 1 a 1 a 2 . . . a n n = a 2 a 1 a 2 . . . a n n = . . . = a n a 1 a 2 . . . a n n {displaystyle {frac {a_{1}}{sqrt{a_{1}a_{2}...a_{n}}}}={frac {a_{2}}{sqrt{a_{1}a_{2}...a_{n}}}}=...={frac {a_{n}}{sqrt{a_{1}a_{2}...a_{n}}}}} 即 a 1 = a 2 = ⋯ = a n {displaystyle a_{1}=a_{2}=cdots =a_{n}} 时等号成立。整理即得： a 1 + a 2 + . . . + a n n ⩾ a 1 a 2 . . . a n n {displaystyle {frac {a_{1}+a_{2}+...+a_{n}}{n}}geqslant {sqrt{a_{1}a_{2}...a_{n}}}} ，当且仅当 a 1 = a 2 = ⋯ = a n {displaystyle a_{1}=a_{2}=cdots =a_{n}} 时等号成立。于是 G n ≤ A n {displaystyle G_{n}leq A_{n}} 得证。利用 G n ≤ A n {displaystyle G_{n}leq A_{n}} ，易证 H n ≤ G n {displaystyle H_{n}leq G_{n}} 。考虑 n {displaystyle n} 个正数 1 a 1 , 1 a 2 , . . . , 1 a n {displaystyle {frac {1}{a_{1}}},{frac {1}{a_{2}}},...,{frac {1}{a_{n}}}} ，有 1 a 1 + 1 a 2 + . . . + 1 a n n ⩾ 1 a 1 ⋅ 1 a 2 ⋅ . . . ⋅ 1 a n n = 1 a 1 a 2 . . . a n n {displaystyle {frac {{frac {1}{a_{1}}}+{frac {1}{a_{2}}}+...+{frac {1}{a_{n}}}}{n}}geqslant {sqrt{{frac {1}{a_{1}}}cdot {frac {1}{a_{2}}}cdot ...cdot {frac {1}{a_{n}}}}}={frac {1}{sqrt{a_{1}a_{2}...a_{n}}}}} ，当且仅当 1 a 1 = 1 a 2 = . . . = 1 a n {displaystyle {frac {1}{a_{1}}}={frac {1}{a_{2}}}=...={frac {1}{a_{n}}}} 即 a 1 = a 2 = ⋯ = a n {displaystyle a_{1}=a_{2}=cdots =a_{n}} 时等号成立。两边取倒数整理得 n 1 a 1 + 1 a 2 + . . . + 1 a n ⩽ a 1 a 2 . . . a n n {displaystyle {frac {n}{{frac {1}{a_{1}}}+{frac {1}{a_{2}}}+...+{frac {1}{a_{n}}}}}leqslant {sqrt{a_{1}a_{2}...a_{n}}}} ，当且仅当 a 1 = a 2 = ⋯ = a n {displaystyle a_{1}=a_{2}=cdots =a_{n}} 时等号成立，即 H n ≤ G n {displaystyle H_{n}leq G_{n}} 。A n ≤ Q n {displaystyle A_{n}leq Q_{n}} 等价于 Q n 2 − A n 2 ≥ 0 {displaystyle Q_{n}^{2}-A_{n}^{2}geq 0} 。事实上， Q n 2 − A n 2 {displaystyle Q_{n}^{2}-A_{n}^{2}} 等于 a 1 , a 2 , … , a n {displaystyle a_{1},a_{2},ldots ,a_{n}} 的方差，通过这个转化可以证出 Q n 2 − A n 2 ≥ 0 {displaystyle Q_{n}^{2}-A_{n}^{2}geq 0} ，证明如下。Q n 2 − A n 2 = Q n 2 − 2 A n 2 + A n 2 = a 1 2 + a 2 2 + … + a n 2 n − 2 a 1 A n + 2 a 2 A n + … + 2 a n A n n + n A n 2 n {displaystyle Q_{n}^{2}-A_{n}^{2}=Q_{n}^{2}-2A_{n}^{2}+A_{n}^{2}={frac {a_{1}^{2}+a_{2}^{2}+ldots +a_{n}^{2}}{n}}-{frac {2a_{1}A_{n}+2a_{2}A_{n}+ldots +2a_{n}A_{n}}{n}}+{frac {nA_{n}^{2}}{n}}}= ( a 1 2 − 2 a 1 A n + A n 2 ) + ( a 2 2 − 2 a 2 A n + A n 2 ) + … + ( a n 2 − 2 a n A n + A n 2 ) n {displaystyle ={frac {(a_{1}^{2}-2a_{1}A_{n}+A_{n}^{2})+(a_{2}^{2}-2a_{2}A_{n}+A_{n}^{2})+ldots +(a_{n}^{2}-2a_{n}A_{n}+A_{n}^{2})}{n}}}= ( a 1 − A n ) 2 + ( a 2 − A n ) 2 + … + ( a n − A n ) 2 n {displaystyle ={frac {(a_{1}-A_{n})^{2}+(a_{2}-A_{n})^{2}+ldots +(a_{n}-A_{n})^{2}}{n}}}⩾ 0 {displaystyle geqslant 0} ，当且仅当 a 1 = a 2 = ⋯ = a n = A n {displaystyle a_{1}=a_{2}=cdots =a_{n}=A_{n}} 时等号成立。利用琴生不等式法也可以很简单地证明均值不等式，同时还有柯西归纳法等方法。