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

  3. 贝努利不等式
    (1) 已知 (x>-1,)(alpha>1)(alpha<0)((1+x)^{alpha} geq 1+alpha mathrm{x})
    (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) cdotsleft(1+x_{n}right)>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})(1, cdots, mathrm{n}) 的一个排列
    (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})

  5. 舒尔(Schur)不等式 已知 (x, y, z geq 0,)(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 xright)^{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),)(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])
    (frac{fleft(x_{1}right)+cdots+fleft(x_{n}right)}{n} geq fleft(frac{x_{1}+cdots+x_{n}}{n}right),) 当且仅当 (x_{1}=cdots=x_{n}) 时取等

  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 aright))
    ((3)left(a b^{2}+b c^{2}+c a^{2}right)-left(a^{2} b+b^{2} c+c^{2} aright)=(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_{i=1}^{n} a_{i}right)^{2}=sum_{i=1}^{n} a_{i}^{2}+2 sum_{i<j} a_{i} a_{j})
    ((8))(Delta mathrm{ABC})(, sum cos ^{2} A+2 cos A cos B cos C=1, sum cot B cot C=1)

  8. 三角横等式

    [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 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 (3 alpha)=3 sin alpha-4sin^3 alpha=4 sin alpha cdot sin left(60^{circ}+alpharight) sin left(60^{circ}-alpharight)
    cos (3 alpha)=4cos^3 alpha-3 cos alpha=4 cos alpha cdot cos left(60^{circ}+alpharight) cos left(60^{circ}-alpharight)
    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}]
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