均值不等式

对正实数 a,ba,b,有

21a+1baba+b2a2+b22当且仅当  a=b  时,等号成立\begin{align*} \dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}\leq \sqrt{ab} \leq\dfrac{a+b}{2} \leq\sqrt{\dfrac{a^2+b^2}{2}}\\ 当且仅当\;a=b\;时,等号成立 \end{align*}

21a+1b\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}} 称作 a,ba,b 的调和平均值

ab\sqrt{ab} 称作 a,ba,b 的几何平均值

a+b2\dfrac{a+b}{2} 称作 a,ba,b 的算术平均值

a2+b22\sqrt{\dfrac{a^2+b^2}{2}} 称作 a,ba,b 的平方平均值

上面的不等式链可简记为“调几算方”

  • 推广
    对正实数 a1,a2,...,ana_{1},a_{2},...,a_{n},有

n1a1+1a2+...+1ana1a2...anna1+a2+....+anna12+a22+....+an2n当且仅当  a1=a2=...an  时,等号成立\dfrac{n}{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+...+\dfrac{1}{a_{n}}}\leq \sqrt[n]{a_{1}a_{2}...a_{n}} \leq \dfrac{a_{1}+a_{2}+....+a_{n}}{n} \leq \sqrt{\dfrac{a_{1}^2+a_{2}^2+....+a_{n}^2}{n}}\\ 当且仅当\;a_{1}=a_{2}=...a_{n}\;时,等号成立

绝对值不等式

aba±ba+b|a|-|b|\leq|a\pm b|\leq|a|+|b|

柯西不等式

对实数 a1,a2,...,an,b1,b2,...,bna_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n},有

(a12+a22+...an2)(b12+b22+...+bn2)(a1b1+a2b2+...+anbn)2  a1b1=a2b2=...anbn  时,等号成立(a_{1}^2+a_{2}^2+...a_{n}^2)(b_{1}^2+b_{2}^2+...+b_{n}^2)\geq(a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n})^2\\[2ex] 当\;\dfrac{a_{1}}{b_{1}}=\dfrac{a_{2}}{b_{2}}=...\dfrac{a_{n}}{b_{n}}\;时,等号成立

积分不等式

1、CauthySchwarz\mathrm{Cauthy-Schwarz} 不等式

[abf(x)g(x) dx]2abf2(x) dxabg2(x) dx \left[ \int_a^b f(x)g(x)\ \mathrm dx \right]^2 \leq \int_a^b f^2(x)\ \mathrm dx\int_a^bg^2(x)\ \mathrm dx

证明:

ab[tf(x)+g(x)]2 dx0 \int_a^b \left[tf(x)+g(x)\right]^2\ \mathrm dx \geq 0

t2abf2(x) dx+2tabf(x)g(x) dx+abg2(x) dx0t^2\int_a^b f^2(x)\ \mathrm dx + 2t\int_a^b f(x)g(x)\ \mathrm dx+\int_a^b g^2(x)\ \mathrm dx \geq 0

Δ0\Delta\leq0 得,

[2abf(x)g(x) dx]24abf2(x)dxabg2(x) dx0 \left[ 2\int_a^bf(x)g(x)\ \mathrm dx \right]^2 - 4\int_a^bf^2(x)dx\int_a^bg^2(x)\ \mathrm dx \leq0

[abf(x)g(x) dx]2abf2(x) dxabg2(x) dx \left[ \int_a^bf(x)g(x)\ \mathrm dx \right]^2 \leq \int_a^bf^2(x)\ \mathrm dx\int_a^bg^2(x)\ \mathrm dx

2、

f(x)f(x)[a,b][a,b] 上非负可积,且 abf(x) dx=1\int_a^bf(x)\ \mathrm dx=1,则

[abf(x)cos(αx) dx]2+[abf(x)sin(αx) dx]21\left[\int_a^bf(x)\cos(\alpha x)\ \mathrm dx\right]^2+\left[\int_a^bf(x)\sin(\alpha x)\ \mathrm dx\right]^2\leq1

证明:

依题意得,f(x)\sqrt{f(x)}f(x)sin(αx)\sqrt{f(x)}\sin(\alpha x)f(x)cos(αx)\sqrt{f(x)}\cos(\alpha x)[a,b][a,b] 上可积,则由 CauthySchwarz\mathrm{Cauthy-Schwarz} 不等式得

[abf(x)cos(αx) dx]2=[abf(x)f(x)cos(αx) dx]2abf(x) dxabf(x)cos2(αx) dx=abf(x)cos2(αx) dx\begin{aligned} \left[\int_a^bf(x)\cos(\alpha x)\ \mathrm dx\right]^2&= \left[\int_a^b\sqrt{f(x)}\cdot\sqrt{f(x)}\cos(\alpha x)\ \mathrm dx\right]^2\\ &\leq\int_a^b f(x)\ \mathrm dx\int_a^bf(x)\cos^2(\alpha x)\ \mathrm dx\\ &=\int_a^bf(x)\cos^2(\alpha x)\ \mathrm dx \end{aligned}

[abf(x)sin(αx)dx]2=[abf(x)f(x)sin(αx) dx]2abf(x) dxabf(x)sin2(αx) dx=abf(x)sin2(αx) dx\begin{aligned} \left[\int_a^bf(x)\sin(\alpha x)\mathrm dx\right]^2&= \left[\int_a^b\sqrt{f(x)}\cdot\sqrt{f(x)}\sin(\alpha x)\ \mathrm dx\right]^2\\ &\leq\int_a^b f(x)\ \mathrm dx\int_a^bf(x)\sin^2(\alpha x)\ \mathrm dx\\ &=\int_a^bf(x)\sin^2(\alpha x)\ \mathrm dx \end{aligned}

两式相加有

[abf(x)cos(αx) dx]2+[abf(x)sin(αx) dx]2abf(x)cos2(αx) dx+abf(x)sin2(αx) dx=abf(x) dx=1\begin{aligned} \left[\int_a^bf(x)\cos(\alpha x)\ \mathrm dx\right]^2+\left[\int_a^bf(x)\sin(\alpha x)\ \mathrm dx\right]^2&\leq \int_a^bf(x)\cos^2(\alpha x)\ \mathrm dx+\int_a^bf(x)\sin^2(\alpha x)\ \mathrm dx\\ &=\int_a^bf(x)\ \mathrm dx=1 \end{aligned}