几个常用的不等式

1.伯努利不等式

命题
h > − 1 , n ∈ N + h>-1, n \in \mathbf{N}_{+} h>1,nN+,则成立不等式
( 1 + h ) n ⩾ 1 + n h (1+h)^{n} \geqslant 1+n h (1+h)n1+nh
其中当 n > 1 n>1 n>1时成立等号的充分必要条件是 h = 0 h=0 h=0

证明:由于 n = 1 n=1 n=1 h = 0 h=0 h=0时不等式明显成立(且其中均成立等号),一下只需讨论 n > 1 n>1 n>1 h ≠ 0 h \neq 0 h=0的情况。
( 1 + h ) n − 1 (1+h)^{n}-1 (1+h)n1做因式分解,就可以得到
( 1 + h ) n − 1 = h [ 1 + ( 1 + h ) + ( 1 + h ) 2 + ⋯ + ( 1 + h ) n − 1 ] (1+h)^{n}-1=h\left[1+(1+h)+(1+h)^{2}+\cdots+(1+h)^{n-1}\right] (1+h)n1=h[1+(1+h)+(1+h)2++(1+h)n1]
h > 0 h>0 h>0时,在右边方括号内从第二项起都大于1,因此就有 ( 1 + h ) n − 1 > n h (1+h)^{n}-1>n h (1+h)n1>nh
− 1 < h < 0 -1<h<0 1<h<0时在上述公式右边方括号中从第二项起都小于1,因此方括号中表达式之和小于 n n n,由于 h < 0 h<0 h<0,因此得到 ( 1 + h ) n − 1 > n h (1+h)^{n}-1>n h (1+h)n1>nh
推广:令 h = B / A h=B/A h=B/A,其中 A > 0 , A + B > 0 A>0, A+B>0 A>0,A+B>0,则条件 1 + h > 0 1+h>0 1+h>0成立,将这个 h h h代入伯努利不等式中,就可以得到下一个不等式:
命题:设有 A > 0 , A + B > 0 , n ∈ N + A>0, A+B>0, n \in \mathbf{N}_{+} A>0,A+B>0,nN+,则成立不等式 ( A + B ) n ⩾ A n + n A n − 1 B (A+B)^{n} \geqslant A^{n}+n A^{n-1} B (A+B)nAn+nAn1B,而且当 n > 1 n>1 n>1时其中成立等号的充分必要条件是 B = 0 B=0 B=0

2.算术平均值-几何平均值不等式

命题:设 a 1 , a 2 , ⋯   , a n a_{1}, a_{2}, \cdots, a_{n} a1,a2,,an n n n个非负实数,则成立不等式:
a 1 + a 2 + ⋯ + a n n ⩾ a 1 a 2 ⋯ a n n \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \geqslant \sqrt[n]{a_{1} a_{2} \cdots a_{n}} na1+a2++anna1a2an
其中等号成立的充分必要条件是 a 1 = a 2 = ⋯ = a n a_{1}=a_{2}=\cdots=a_{n} a1=a2==an

证1:一开始可以看出,如果在 a 1 , a 2 , ⋯   , a n a_{1}, a_{2}, \cdots, a_{n} a1,a2,,an中出现0,则不等式已经成立,又可以看出,这时成立等号的充分必要条件是其中每个数为0,因此下面只需要对 a 1 , a 2 , ⋯   , a n a_{1}, a_{2}, \cdots, a_{n} a1,a2,,an n n n个正数的情况来证明就够了。
应用数学归纳法,在 n = 1 n=1 n=1时结论是平凡的,在 n = 2 n=2 n=2时的结论是中学数学已包含的内容,现设 n = k n=k n=k时不等式已成立,然后讨论 n = k + 1 n=k+1 n=k+1,将 k + 1 k+1 k+1个正数 a 1 , a 2 , ⋯   , a n a_{1}, a_{2}, \cdots, a_{n} a1,a2,,an的算术平均值分解如下:
a 1 + a 2 + ⋯ + a k + 1 k + 1 = a 1 + a 2 + ⋯ + a k k + k a k + 1 − ( a 1 + a 2 + ⋯ + a k ) k ( k + 1 ) \frac{a_{1}+a_{2}+\cdots+a_{k+1}}{k+1}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}+\frac{k a_{k+1}-\left(a_{1}+a_{2}+\cdots+a_{k}\right)}{k(k+1)} k+1a1+a2++ak+1=ka1+a2++ak+k(k+1)kak+1(a1+a2++ak)
然后将上式右边的两项分别记为 A A A B B B,这时条件 A > 0 , A + B > 0 A>0, A+B>0 A>0,A+B>0满足,因此就可以应用伯努利不等式的推广不等式进行如下计算:
( a 1 + a 2 + ⋯ + a k + 1 k + 1 ) k + 1 = ( A + B ) k + 1 ⩾ A k + 1 + ( k + 1 ) A k B = A k ( A + ( k + 1 ) B ) = ( a 1 + a 2 + ⋯ + a k k ) k ⋅ a k + 1 ⩾ a 1 a 2 ⋯ a k a k + 1 \begin{aligned} \left(\frac{a_{1}+a_{2}+\cdots+a_{k+1}}{k+1}\right)^{k+1} &=(A+B)^{k+1} \geqslant A^{k+1}+(k+1) A^{k} B \\ &=A^{k}(A+(k+1) B) \\ &=\left(\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}\right)^{k} \cdot a_{k+1} \\ & \geqslant a_{1} a_{2} \cdots a_{k} a_{k+1} \end{aligned} (k+1a1+a2++ak+1)k+1=(A+B)k+1Ak+1+(k+1)AkB=Ak(A+(k+1)B)=(ka1+a2++ak)kak+1a1a2akak+1
在不等式中成立等号的条件也可以用数学归纳法得到,在 n = 1 n=1 n=1时已成立,设在 n = k n=k n=k时结论为真,则在 n = k + 1 n=k+1 n=k+1时可从上述推导看出等号成立的条件是 k a k + 1 = a 1 + a 2 + ⋯ + a k k a_{k+1}=a_{1}+a_{2}+\cdots+a_{k} kak+1=a1+a2++ak a 1 = a 2 = ⋯ = a k a_{1}=a_{2}=\cdots=a_{k} a1=a2==ak
也即是 a 1 = a 2 = ⋯ = a k = a k + 1 a_{1}=a_{2}=\cdots=a_{k}=a_{k+1} a1=a2==ak=ak+1

证2:采用拉格朗日乘数法,令 a 1 a 2 ⋯ a n = a a_{1} a_{2} \cdots a_{n}=a a1a2an=a,并做辅助函数
F = a 1 + a 2 + ⋯ + a n + λ ( a 1 a 2 ⋯ a n − a ) F=a_{1}+a_{2}+\dots+a_{n}+\lambda\left(a_{1} a_{2} \cdots a_{n}-a\right) F=a1+a2++an+λ(a1a2ana)

{ F a 1 ′ = 1 + λ a 2 a 3 ⋯ a n = 0 F a 2 ′ = 1 + λ a 1 a 3 ⋯ a n = 0 ⋯ ⋯ F a n ′ = 1 + λ a 1 a 2 ⋯ a n − 1 = 0 F λ ′ = a 1 a 2 ⋯ a n − a = 0 \left\{\begin{array}{l}{F_{a_{1}}^{\prime}=1+\lambda a_{2} a_{3} \cdots a_{n}=0} \\ {F_{a_{2}}^{\prime}=1+\lambda a_{1} a_{3} \cdots a_{n}=0} \\ {\cdots \cdots} \\ {F_{a_{n}}^{\prime}=1+\lambda a_{1} a_{2} \cdots a_{n-1}=0} \\ {F_{\lambda}^{\prime}=a_{1} a_{2} \cdots a_{n}-a=0}\end{array}\right. Fa1=1+λa2a3an=0Fa2=1+λa1a3an=0Fan=1+λa1a2an1=0Fλ=a1a2ana=0
a 1 = a 2 = ⋯ = a n = a n a_{1}=a_{2}=\dots=a_{n}=\sqrt[n]{a} a1=a2==an=na
由于 a 1 , a 2 , ⋯   , a n > 0 a_{1}, a_{2}, \cdots, a_{n}>0 a1,a2,,an>0,所以 a 1 + a 2 + ⋯ + a n a_{1}+a_{2}+\dots+a_{n} a1+a2++an无最大值,其最小值为
( a 1 + a 2 + ⋯ + a n ) min ⁡ = n a n \left(a_{1}+a_{2}+\cdots+a_{n}\right)_{\min }=n \sqrt[n]{a} (a1+a2++an)min=nna

a 1 + a 2 + ⋯ + a n ⩾ n a n = n a 1 a 2 ⋯ a n n a_{1}+a_{2}+\dots+a_{n} \geqslant n \sqrt[n]{a}=n \sqrt[n]{a_{1} a_{2} \cdots a_{n}} a1+a2++annna =nna1a2an
得证

3.几何平均值-调和平均值不等式

命题:若 a k > 0 , k = 1 , 2 , ⋯   , n a_{k}>0, k=1,2, \cdots, n ak>0,k=1,2,,n,则有
( ∏ k = 1 n a k ) 1 n ⩾ n ∑ k = 1 n 1 a k \left(\prod_{k=1}^{n} a_{k}\right)^{\frac{1}{n}} \geqslant \frac{n}{\sum_{k=1}^{n} \frac{1}{a_{k}}} (k=1nak)n1k=1nak1n

证明:首先对数函数具有凸性,要证算术平均值-几何平均值不等式:
1 n ∑ k = 1 n a k ≥ ∏ k = 1 n a k \frac{1}{n} \sum_{k=1}^{n} a_{k} \geq \sqrt{\prod_{k=1}^{n} a_{k}} n1k=1nakk=1nak
两边取对数,即证:
ln ⁡ ( 1 n ∑ k = 1 n a k ) ≥ 1 n ∑ k = 1 n ln ⁡ a k \ln \left(\frac{1}{n} \sum_{k=1}^{n} a_{k}\right) \geq \frac{1}{n} \sum_{k=1}^{n} \ln a_{k} ln(n1k=1nak)n1k=1nlnak
对于几何平均值-调和平均值不等式而言:
∏ k = 1 n a k n ≥ n ∑ k = 1 n 1 a k \sqrt[n]{\prod_{k=1}^{n} a_{k}} \geq \frac{n}{\sum_{k=1}^{n} \frac{1}{a_{k}}} nk=1nak k=1nak1n
先取倒数,得:
∏ k = 1 n 1 a k n ≤ 1 n ∑ k = 1 n 1 a k \sqrt[n]{\prod_{k=1}^{n} \frac{1}{a_{k}}} \leq \frac{1}{n} \sum_{k=1}^{n} \frac{1}{a_{k}} nk=1nak1 n1k=1nak1
再取对数:
1 n ∑ k = 1 n ln ⁡ 1 a k ≤ ln ⁡ ( 1 n ∑ k = 1 n 1 a k ) \frac{1}{n} \sum_{k=1}^{n} \ln \frac{1}{a_{k}} \leq \ln \left(\frac{1}{n} \sum_{k=1}^{n} \frac{1}{a_{k}}\right) n1k=1nlnak1ln(n1k=1nak1)
证毕。

4.柯西不等式

命题:对实数 a 1 , a 2 , ⋯   , a n a_{1}, a_{2}, \cdots, a_{n} a1,a2,,an b 1 , b 2 , ⋯   , b n b_{1}, b_{2}, \cdots, b_{n} b1,b2,,bn成立
∣ ∑ i = 1 n a i b i ∣ ⩽ ∑ i = 1 n a i 2 ∑ i = 1 n b i 2 \left|\sum_{i=1}^{n} a_{i} b_{i}\right| \leqslant \sqrt{\sum_{i=1}^{n} a_{i}^{2}} \sqrt{\sum_{i=1}^{n} b_{i}^{2}} i=1naibii=1nai2 i=1nbi2

证明:引入变量 λ \lambda λ,写出如下的非负二次三项式
0 ⩽ ∑ i = 1 n ( λ a i − b i ) 2 = λ 2 ∑ i = 1 n a i 2 − 2 λ ∑ i = 1 n a i b i + ∑ i = 1 n b i 2 0 \leqslant \sum_{i=1}^{n}\left(\lambda a_{i}-b_{i}\right)^{2}=\lambda^{2} \sum_{i=1}^{n} a_{i}^{2}-2 \lambda \sum_{i=1}^{n} a_{i} b_{i}+\sum_{i=1}^{n} b_{i}^{2} 0i=1n(λaibi)2=λ2i=1nai22λi=1naibi+i=1nbi2
如果 a 1 , a 2 , ⋯   , a n a_{1}, a_{2}, \cdots, a_{n} a1,a2,,an全为0,可以发现柯西不等式已成立。否则, λ 2 \lambda^{2} λ2项的系数不会是0,因此它的判别式非正,这就导致
( ∑ i = 1 n a i b i ) 2 ⩽ ( ∑ i = 1 n a i 2 ) ⋅ ( ∑ i = 1 n b i 2 ) \left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} \leqslant\left(\sum_{i=1}^{n} a_{i}^{2}\right) \cdot\left(\sum_{i=1}^{n} b_{i}^{2}\right) (i=1naibi)2(i=1nai2)(i=1nbi2)
两边开方,就得到所要求证的不等式。
补充:在柯西不得鞥是中等号成立的充分必要条件是两个序列 { a i } 1 ⩽ i ⩽ n \left\{a_{i}\right\}_{1 \leqslant i \leqslant n} {ai}1in { b i } 1 ⩽ i ⩽ n \left\{b_{i}\right\}_{1 \leqslant i \leqslant n} {bi}1in成比例。

5.其它不等式
  1. x > 0 , y > 0 , p > 0 , q > 0 , 1 p + 1 q = 1 x>0, y>0, p>0, q>0, \frac{1}{p}+\frac{1}{q}=1 x>0,y>0,p>0,q>0,p1+q1=1,则有 x y ⩽ x p p + y q q x y \leqslant \frac{x^{p}}{p}+\frac{y^{q}}{q} xypxp+qyq

证明:在 x y ⩽ x p p + y q q x y \leqslant \frac{x^{p}}{p}+\frac{y^{q}}{q} xypxp+qyq两边取对数,得到 ln ⁡ ( x y ) ⩽ ln ⁡ ( x p p + y q q ) \ln (x y) \leqslant \ln \left(\frac{x^{p}}{p}+\frac{y^{q}}{q}\right) ln(xy)ln(pxp+qyq)
由于 x > 0 x>0 x>0,令 f ( x ) = ln ⁡ x ⇒ f ′ ′ ( x ) = − 1 x 2 < 0 ⇒ f ( x ) f(x)=\ln x \Rightarrow f^{\prime \prime}(x)=-\frac{1}{x^{2}}<0 \Rightarrow f(x) f(x)=lnxf(x)=x21<0f(x)的图像是凸的,,由凸凹性定义得:
f [ λ x 1 + ( 1 − λ ) y 1 ] ⩾ λ f ( x 1 ) + ( 1 − λ ) f ( y 1 ) f\left[\lambda x_{1}+(1-\lambda) y_{1}\right] \geqslant \lambda f\left(x_{1}\right)+(1-\lambda) f\left(y_{1}\right) f[λx1+(1λ)y1]λf(x1)+(1λ)f(y1)
在上式中,令 λ = 1 p , x 1 = x p , y 1 = y q \lambda=\frac{1}{p}, x_{1}=x^{p}, y_{1}=y^{q} λ=p1,x1=xp,y1=yq,则 1 − λ = 1 − 1 p = 1 q 1-\lambda=1-\frac{1}{p}=\frac{1}{q} 1λ=1p1=q1,于是就得到
ln ⁡ ( x p p + y q q ) ⩾ 1 p f ( x p ) + 1 q f ( y q ) = ln ⁡ ( x y ) \ln \left(\frac{x^{p}}{p}+\frac{y^{q}}{q}\right) \geqslant \frac{1}{p} f\left(x^{p}\right)+\frac{1}{q} f\left(y^{q}\right)=\ln (x y) ln(pxp+qyq)p1f(xp)+q1f(yq)=ln(xy)
即得:
x y ⩽ x ρ p + y y q x y \leqslant \frac{x^{\rho}}{p}+\frac{y^{y}}{q} xypxρ+qyy

  1. f ( x ) f(x) f(x) [ a , , b ] [a,,b] [a,,b] p p p次方可积, g ( x ) g(x) g(x) [ a , , b ] [a,,b] [a,,b] q q q次方可积,则
    ∣ ∫ a b f ( x ) ⋅ g ( x ) d x ∣ ⩽ [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 p ⋅ [ ∫ a b ∣ g ( x ) ∣ q d x ] 1 q \left|\int_{a}^{b} f(x) \cdot g(x) \mathrm{d} x\right| \leqslant\left[\int_{a}^{b}|f(x)|^{p} \mathrm{d} x\right]^{\frac{1}{p}} \cdot\left[\int_{a}^{b}|g(x)|^{q} \mathrm{d} x\right]^{\frac{1}{q}} abf(x)g(x)dx[abf(x)pdx]p1[abg(x)qdx]q1
    其中 p > 1 , 1 p + 1 q = 1 p>1, \frac{1}{p}+\frac{1}{q}=1 p>1,p1+q1=1

 

证明:设 A = ∣ f ( x ) ∣ [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 p , B = ∣ g ( x ) ∣ [ ∫ a b ∣ g ( x ) ∣ q d x ] 1 q A=\frac{|f(x)|}{\left[\int_{a}^{b}|f(x)|^{p} \mathrm{d} x\right]^{\frac{1}{p}}}, B=\frac{|g(x)|}{\left[\int_{a}^{b}|g(x)|^{q} \mathrm{d} x\right]^{\frac{1}{q}}} A=[abf(x)pdx]p1f(x),B=[abg(x)qdx]q1g(x),利用Young不等式得:
∣ f ( x ) ∣ [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 p ⋅ ∣ g ( x ) ∣ [ ∫ a b ∣ g ( x ) ∣ q d x ] 1 q ⩽ 1 p ∣ f ( x ) ∣ p ∫ a b ∣ f ( x ) ∣ p d x + 1 q ∣ g ( x ) ∣ q ∫ a b ∣ g ( x ) ∣ s d x \frac{|f(x)|}{\left[\int_{a}^{b}|f(x)|^{p} \mathrm{d} x\right]^{\frac{1}{p}}} \cdot \frac{|g(x)|}{\left[\int_{a}^{b}|g(x)|^{q} d x\right]^{\frac{1}{q}}} \leqslant \frac{1}{p} \frac{|f(x)|^{p}}{\int_{a}^{b}|f(x)|^{p} \mathrm{d} x}+\frac{1}{q} \frac{|g(x)|^{q}}{\int_{a}^{b}|g(x)|^{s} d x} [abf(x)pdx]p1f(x)[abg(x)qdx]q1g(x)p1abf(x)pdxf(x)p+q1abg(x)sdxg(x)q
于是
∣ f ( x ) ∣ ∣ g ( x ) ∣ ⩽ 1 p ∣ f ( x ) ∣ p [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 − 1 p [ ∫ a b ∣ g ( x ) ∣ q d x ] 1 q + 1 q ∣ g ( x ) ∣ q [ ∫ a b ∣ g ( x ) ∣ q d x ] 1 − 1 q [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 p = 1 p ∣ f ( x ) ∣ ⋅ [ ∫ a b ∣ g ( x ) ∣ q d x ] 1 q [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 q + 1 q ∣ g ( x ) ∣ q [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 p [ ∫ a b ∣ g ( x ) ∣ r d x ] 1 p \begin{aligned}|f(x)||g(x)| \leqslant \frac{1}{p} & \frac{|f(x)|^{p}}{\left[\int_{a}^{b}|f(x)|^{p} \mathrm{d} x\right]^{1-\frac{1}{p}}}\left[\int_{a}^{b}|g(x)|^{q} \mathrm{d} x\right]^{\frac{1}{q}}+\\ & \frac{1}{q} \frac{|g(x)|^{q}}{\left[\int_{a}^{b}|g(x)|^{q} d x\right]^{1-\frac{1}{q}}}\left[\int_{a}^{b}|f(x)|^{p} \mathrm{d} x\right]^{\frac{1}{p}} \\ & =\frac{1}{p}|f(x)| \cdot \frac{\left[\int_{a}^{b}|g(x)|^{q} d x\right]^{\frac{1}{q}}}{\left[\int_{a}^{b}|f(x)|^{p} d x\right]^{\frac{1}{q}}}+\frac{1}{q}|g(x)|^{q} \frac{\left[\int_{a}^{b}|f(x)|^{p} d x\right]^{\frac{1}{p}}}{\left[\int_{a}^{b}|g(x)|^{r} d x\right]^{\frac{1}{p}}} \end{aligned} f(x)g(x)p1[abf(x)pdx]1p1f(x)p[abg(x)qdx]q1+q1[abg(x)qdx]1q1g(x)q[abf(x)pdx]p1=p1f(x)[abf(x)pdx]q1[abg(x)qdx]q1+q1g(x)q[abg(x)rdx]p1[abf(x)pdx]p1
对上式两边在 [ a , b ] [a,b] [a,b]上关于 x x x积分,得:
∫ a b ∣ f ( x ) g ( x ) ∣ d x ⩽ 1 p ∫ a b ∣ f ( x ) ∣ p d x [ ∫ a b ∣ g ( x ) ∣ q d x ] 1 q [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 q + 1 q ∫ a b ∣ g ( x ) ∣ q d x [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 p [ ∫ a b ∣ g ( x ) ∣ q d x ] − 1 p = ( 1 p + 1 q ) [ ∫ a b ∣ f ( x ) ∣ p d x ] 1 p ⋅ [ ∫ a b ∣ g ( x ) ∣ q d x ] 1 q \begin{aligned} \int_{a}^{b}|f(x) g(x)| d x & \leqslant \frac{1}{p} \int_{a}^{b}|f(x)|^{p} d x \frac{\left[\int_{a}^{b}|g(x)|^{q} d x\right]^{\frac{1}{q}}}{\left[\int_{a}^{b}|f(x)|^{p} d x\right]^{\frac{1}{q}}}+\frac{1}{q} \int_{a}^{b}|g(x)|^{q} d x \frac{\left[\int_{a}^{b}|f(x)|^{p} d x\right]^{\frac{1}{p}}}{\left[\int_{a}^{b}|g(x)|^{q} d x\right]^{-\frac{1}{p}}} \\ &=\left(\frac{1}{p}+\frac{1}{q}\right)\left[\int_{a}^{b}|f(x)|^{p} d x\right]^{\frac{1}{p}} \cdot\left[\int_{a}^{b}|g(x)|^{q} d x\right]^{\frac{1}{q}} \end{aligned} abf(x)g(x)dxp1abf(x)pdx[abf(x)pdx]q1[abg(x)qdx]q1+q1abg(x)qdx[abg(x)qdx]p1[abf(x)pdx]p1=(p1+q1)[abf(x)pdx]p1[abg(x)qdx]q1
又由 ∣ ∫ a b f ( x ) ⋅ g ( x ) d x ∣ ⩽ ∫ a b ∣ f ( x ) g ( x ) ∣ d x \left|\int_{a}^{b} f(x) \cdot g(x) \mathrm{d} x\right| \leqslant \int_{a}^{b}|f(x) g(x)| \mathrm{d} x abf(x)g(x)dxabf(x)g(x)dx,命题得证。

  1. arctan ⁡ x ⩽ x ⩽ arcsin ⁡ x ( 0 ⩽ x ⩽ 1 ) \arctan x \leqslant x \leqslant \arcsin x(0 \leqslant x \leqslant 1) arctanxxarcsinx(0x1)
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