数学竞赛常用结论
想不到吧，我转竞赛了QWQ
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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
\( \frac{n}{\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}} \leq \sqrt[n]{a_{1} \cdots a_{n}} \leq \frac{a_{1}+\cdots+a_{n}}{n} \leq \sqrt{\frac{a_{1}^{2}+\cdots+a_{n}^{2}}{n}} \)
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柯西不等式、拉格朗日恒等式与赫尔德不等式
\(\left(a_{1}^{2}+\cdots+a_{n}^{2}\right)\left(b_{1}^{2}+\cdots+b_{n}^{2}\right) \geq\left(a_{1} b_{1}+\cdots+a_{n} b_{n}\right)^{2},\)当且仅当\(\frac{a_{1}}{b_{1}}=\cdots=\frac{a_{n}}{b_{n}}\) 时取等
\[\left(\sum\limits_{i=1}^{n} a_{i}^{2}\right)\left(\sum\limits_{i=1}^{n} b_{i}^{2}\right)=\left(\sum\limits_{i=1}^{n} a_{i} b_{i}\right)^{2}+\sum\limits_{1 \leq i\(\left(a_{1}^{3}+\cdots+a_{n}^{3}\right)\left(b_{1}^{3}+\cdots+b_{n}^{3}\right)\left(c_{1}^{3}+\cdots+c_{n}^{3}\right) \geq\left(a_{1} b_{1} c_{1}+\cdots+a_{n} b_{n} c_{n}\right)^{3},\) 各项为正
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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, \(\left(1+x_{1}\right) \cdots\left(1+x_{n}\right)>1+x_{1}+\cdots+x_{n}\)
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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\(\begin{array}{l}a_{1} b_{1}+\cdots+a_{n} b_{n} \geq a_{1} b_{i_{1}}+\cdots+a_{n} b_{i_{n}} \geq a_{1} b_{n}+\cdots+a_{n} b_{1} \\ a_{1} b_{1}+\cdots+a_{n} b_{n} \geq \frac{1}{n}\left(a_{1}+\cdots+a_{n}\right)\left(b_{1}+\cdots+b_{n}\right) \geq a_{1} b_{n}+\cdots+a_{n} b_{1}\end{array}\)
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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)\(\left(\sum x\right)^{3}-4 \sum x \sum x y+9 x y z \geq 0\)
(3) $\prod(x+y-z) \leq xyz $
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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 \(\frac{f\left(x_{1}\right)+\cdots+f\left(x_{n}\right)}{n} \geq f\left(\frac{x_{1}+\cdots+x_{n}}{n}\right),\) 当且仅当 \(x_{1}=\cdots=x_{n}\) 时取等
(3)更强的结论：若\(\lambda_i\ge 0,\sum\limits_{i=1}^n\lambda_i=1\)，则有\(\lambda_1f(x_1)+\cdots +\lambda_nf(x_n)\le f(\lambda_1x_1+\cdots+\lambda_nx_n)\)
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恒等式
\((1)(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
\((2) a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\)
\((3)\left(a b^{2}+b c^{2}+c a^{2}\right)-\left(a^{2} b+b^{2} c+c^{2} a\right)=(a-b)(b-c)(c-a)\)
\((4)(a+b+c)(a b+b c+c a)=(a+b)(b+c)(c+a)+a b c\)
\((5)(a+b)(a+c)=a^{2}+a b+b c+c a\)
\((6)(a+c)(b+d)=a b+b c+c d+d a\)
\((7)\left(\sum\limits_{i=1}^{n} a_{i}\right)^{2}=\sum\limits_{i=1}^{n} a_{i}^{2}+2 \sum\limits_{i
\((8)\)在 \(\Delta \mathrm{ABC}\) 中 \(, \sum \cos ^{2} A+2 \cos A \cos B \cos C=1, \sum \cot B \cot C=1\)
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三角横等式
\[\begin{array}{l} \sin \alpha+\sin \beta=2 \sin \frac {\alpha+\beta}2 \cos \frac {\alpha-\beta}2 \\ \sin \alpha-\sin \beta=2 \cos \frac {\alpha+\beta}2 \sin \frac {\alpha-\beta}2 \\ \cos \alpha+\cos \beta=2 \cos \frac {\alpha+\beta}2 \cos \frac {\alpha-\beta}2 \\ \cos \alpha-\cos \beta=-2 \sin \frac {\alpha+\beta}2 \sin \frac {\alpha-\beta}2 \end{array}\]\[\begin{array}{l} \sin \alpha \cdot \cos \beta=\frac 12[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \\ \cos \alpha \cdot \sin \beta=\frac 12[\sin (\alpha+\beta)-\sin (\alpha-\beta)] \\ \cos \alpha \cdot \cos \beta=\frac 12[\cos (\alpha+\beta)+\cos (\alpha-\beta)] \\ \sin \alpha \cdot \sin \beta=-\frac 12[\cos (\alpha+\beta)-\cos (\alpha-\beta)] \end{array}\]\[\begin{array}{l} \sin (3 \alpha)=3 \sin \alpha-4\sin^3 \alpha=4 \sin \alpha \cdot \sin \left(60^{\circ}+\alpha\right) \sin \left(60^{\circ}-\alpha\right) \\ \cos (3 \alpha)=4\cos^3 \alpha-3 \cos \alpha=4 \cos \alpha \cdot \cos \left(60^{\circ}+\alpha\right) \cos \left(60^{\circ}-\alpha\right) \\ \end{array}\]\[\begin{aligned} &\sin \alpha=\frac{2 \tan \frac{\alpha}{2}}{1+\tan ^{2} \frac{\alpha}{2}}\\ &\cos \alpha=\frac{1-\tan ^{2} \frac{\alpha}{2}}{1+\tan ^{2} \frac{\alpha}{2}} \end{aligned}\] -
整系数多项式
既约分数\(x_0=\frac pq\)是整系数多项式\(f(x)=\sum\limits_{i=1}^na_ix_i=0\)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
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Abel求和
\[\text {若 } B_{n}=\sum\limits_{i=1}^{n} b_{i} \text { 则 } S=\sum\limits_{i=1}^{n} a_{i} b_{i}=B_{n} a_{n}-\sum\limits_{i=1}^{n-1} B_{i}\left(a_{i+1}-a_{i}\right) \]