数学竞赛常用结论

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1. 均值不等式:若 $ a_{1}, cdots, a_{n}>0,$ 则以下不等式成立，当且仅当 (a_{1}=cdots=a_{n}) 时取等

   ( frac{n}{frac{1}{a_{1}}+cdots+frac{1}{a_{n}}} leq sqrt[n]{a_{1} cdots a_{n}} leq frac{a_{1}+cdots+a_{n}}{n} leq sqrt{frac{a_{1}^{2}+cdots+a_{n}^{2}}{n}} )

2. 柯西不等式、拉格朗日恒等式与赫尔德不等式

(left(a_{1}^{2}+cdots+a_{n}^{2} ight)left(b_{1}^{2}+cdots+b_{n}^{2} ight) geqleft(a_{1} b_{1}+cdots+a_{n} b_{n} ight)^{2},)当且仅当(frac{a_{1}}{b_{1}}=cdots=frac{a_{n}}{b_{n}}) 时取等

[left(sumlimits_{i=1}^{n} a_{i}^{2} ight)left(sumlimits_{i=1}^{n} b_{i}^{2} ight)=left(sumlimits_{i=1}^{n} a_{i} b_{i} ight)^{2}+sumlimits_{1 leq i<j leq n}left(a_{i} b_{j}-a_{j} b_{i} ight)^{2} ]

(left(a_{1}^{3}+cdots+a_{n}^{3} ight)left(b_{1}^{3}+cdots+b_{n}^{3} ight)left(c_{1}^{3}+cdots+c_{n}^{3} ight) geqleft(a_{1} b_{1} c_{1}+cdots+a_{n} b_{n} c_{n} ight)^{3},) 各项为正

3. 贝努利不等式

   (1) 已知 (x>-1,) 当 (alpha>1) 或 (alpha<0) 时((1+x)^{alpha} geq 1+alpha mathrm{x})

   当 (0<alpha<1) 时, ((1+x)^{alpha} leq 1+alpha mathrm{x},) 当且仅当 (x=0) 时取等

   (2) 当(x_{1}, cdots, x_{n}>-1) 且同号时, (left(1+x_{1} ight) cdotsleft(1+x_{n} ight)>1+x_{1}+cdots+x_{n})

4. 排序不等式与切比雪夫不等式

   已知 (a_{1} leq cdots leq a_{n}, b_{1} leq cdots leq b_{n}, i_{1}, cdots, i_{n}) 为 (1, cdots, mathrm{n}) 的一个排列

   则(egin{array}{l}a_{1} b_{1}+cdots+a_{n} b_{n} geq a_{1} b_{i_{1}}+cdots+a_{n} b_{i_{n}} geq a_{1} b_{n}+cdots+a_{n} b_{1} \ a_{1} b_{1}+cdots+a_{n} b_{n} geq frac{1}{n}left(a_{1}+cdots+a_{n} ight)left(b_{1}+cdots+b_{n} ight) geq a_{1} b_{n}+cdots+a_{n} b_{1}end{array})

5. 舒尔(Schur)不等式 已知 (x, y, z geq 0,) 有 (sum x^{r}(x-y)(x-z) geq 0, r=1) 为标准形式,变形

   (1) (sum x^{3}-sum x^{2}(y+z)+3 x y z geq 0)

   (2)(left(sum x ight)^{3}-4 sum x sum x y+9 x y z geq 0)

   (3) $prod(x+y-z) leq xyz $

6. 凸函数与琴生不等式

   (1)若(f'' (x)>0(a leq x leq b),) 则 (f(x)) 在区间 ([a, b]) 上是下凸函数

   (2)若 (f(x)) 满足(1),则对任意 (x_{1}, cdots, x_{n} in[a, b])

   有 (frac{fleft(x_{1} ight)+cdots+fleft(x_{n} ight)}{n} geq fleft(frac{x_{1}+cdots+x_{n}}{n} ight),) 当且仅当 (x_{1}=cdots=x_{n}) 时取等

   (3)更强的结论：若(lambda_ige 0,sumlimits_{i=1}^nlambda_i=1)，则有(lambda_1f(x_1)+cdots +lambda_nf(x_n)le f(lambda_1x_1+cdots+lambda_nx_n))

7. 恒等式

   ((1)(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a)

   ((2) a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)left(a^{2}+b^{2}+c^{2}-a b-b c-c a ight))

   ((3)left(a b^{2}+b c^{2}+c a^{2} ight)-left(a^{2} b+b^{2} c+c^{2} a ight)=(a-b)(b-c)(c-a))

   ((4)(a+b+c)(a b+b c+c a)=(a+b)(b+c)(c+a)+a b c)

   ((5)(a+b)(a+c)=a^{2}+a b+b c+c a)

   ((6)(a+c)(b+d)=a b+b c+c d+d a)

   ((7)left(sumlimits_{i=1}^{n} a_{i} ight)^{2}=sumlimits_{i=1}^{n} a_{i}^{2}+2 sumlimits_{i<j} a_{i} a_{j})

   ((8))在 (Delta mathrm{ABC}) 中 (, sum cos ^{2} A+2 cos A cos B cos C=1, sum cot B cot C=1)

8. 三角横等式

   [egin{array}{l} sin alpha+sin eta=2 sin frac {alpha+eta}2 cos frac {alpha-eta}2 \ sin alpha-sin eta=2 cos frac {alpha+eta}2 sin frac {alpha-eta}2 \ cos alpha+cos eta=2 cos frac {alpha+eta}2 cos frac {alpha-eta}2 \ cos alpha-cos eta=-2 sin frac {alpha+eta}2 sin frac {alpha-eta}2 end{array}]
   [egin{array}{l} sin alpha cdot cos eta=frac 12[sin (alpha+eta)+sin (alpha-eta)] \ cos alpha cdot sin eta=frac 12[sin (alpha+eta)-sin (alpha-eta)] \ cos alpha cdot cos eta=frac 12[cos (alpha+eta)+cos (alpha-eta)] \ sin alpha cdot sin eta=-frac 12[cos (alpha+eta)-cos (alpha-eta)] end{array}]
   [egin{array}{l} sin (3 alpha)=3 sin alpha-4sin^3 alpha=4 sin alpha cdot sin left(60^{circ}+alpha ight) sin left(60^{circ}-alpha ight) \ cos (3 alpha)=4cos^3 alpha-3 cos alpha=4 cos alpha cdot cos left(60^{circ}+alpha ight) cos left(60^{circ}-alpha ight) \ end{array}]
   [egin{aligned} &sin alpha=frac{2 an frac{alpha}{2}}{1+ an ^{2} frac{alpha}{2}}\ &cos alpha=frac{1- an ^{2} frac{alpha}{2}}{1+ an ^{2} frac{alpha}{2}} end{aligned}]

9. 整系数多项式

   既约分数(x_0=frac pq)是整系数多项式(f(x)=sumlimits_{i=1}^na_ix_i=0)的有理数解，则

   (1) (p mid a_{0}) 且 (q mid a_{n})

   (2) (f(x)) 除以 (x-frac{q}{p}) 所得的商的各项系数必为(p)的倍数

10. Abel求和

    [ ext {若 } B_{n}=sumlimits_{i=1}^{n} b_{i} ext { 则 } S=sumlimits_{i=1}^{n} a_{i} b_{i}=B_{n} a_{n}-sumlimits_{i=1}^{n-1} B_{i}left(a_{i+1}-a_{i} ight) ]