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发现了一个不等式

jjason20250630
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命题如下：
对于a1≥a2≥...≥ana_1\geq a_2 \geq ...\geq a_n , b1≤b2≤...≤bnb_1\leq b_2\leq...\le b_n (ai,bi∈R+)(a_i,b_i\in R^+) , 有：

∑i=1naibi≥n∑i=1nai∑i=1nbi\sum_{i=1}^{n} \frac{a_i}{b_i} \geq n\frac{\sum_{i=1}^{n}a_i}{\sum_{i=1}^{n}b_i}

证明如下：
首先证明二元的形式：

a1b1+a2b2≥2×a1+a2b1+b2\frac{a_1}{b_1}+\frac{a_2}{b_2}\geq 2\times\frac{a_1+a_2}{b_1+b_2}

  ⟺  a1b2(b1+b2)+a2b1(b1+b2)≥2b1b2(a1+a2)\iff a_1b_2(b_1+b_2)+a_2b_1(b_1+b_2)\geq2b_1b_2(a_1+a_2)
  ⟺  a1b22+a2b12≥a1b1b2+a2b1b2\iff a_1b_2^2 + a_2b_1^2\geq a_1b_1b_2+a_2b_1b_2
  ⟺  a1b2(b2−b1)≥a2b1(b2−b1)\iff a_1b_2(b_2-b_1)\geq a_2b_1(b_2-b_1)
  ⟺  a1b2≥a2b1\iff a_1b_2\geq a_2b_1
即证
考虑归纳假设：
∀a1≥a2≥...≥an−1\forall a_1\geq a_2 \geq ...\geq a_{n-1} , b1≤b2≤...≤bn−1b_1\leq b_2\leq...\le b_{n-1} (ai,bi∈R+)(a_i,b_i\in R^+)
∑i=1n−1aibi≥(n−1)∑i=1n−1ai∑i=1n−1bi\sum_{i=1}^{n-1} \frac{a_i}{b_i} \geq (n-1)\frac{\sum_{i=1}^{n-1}a_i}{\sum_{i=1}^{n-1}b_i}
则对于an≤an−1,bn≥bn−1a_n\leq a_{n-1},b_n\geq b_{n-1}
有∑i=1naibi=∑i=1n−1aibi+anbn\sum_{i=1}^{n} \frac{a_i}{b_i}=\sum_{i=1}^{n-1} \frac{a_i}{b_i}+\frac{a_n}{b_n}

n∑i=1nai∑i=1nbi=n(∑i=1n−1ai)+an(∑i=1n−1bi)+bn≤n2(∑i=1n−1ai∑i=1n−1bi+anbn)①≤n2(∑i=1n−1aibin−1+anbn)n\frac{\sum_{i=1}^{n}a_i}{\sum_{i=1}^{n}b_i}=n\frac{(\sum_{i=1}^{n-1}a_i)+a_n}{(\sum_{i=1}^{n-1}b_i)+b_n}\leq \frac{n}{2}(\frac{\sum_{i=1}^{n-1}a_i}{\sum_{i=1}^{n-1}b_i}+\frac{a_n}{b_n}){}^{①}\leq \frac{n}{2}(\frac{\sum_{i=1}^{n-1} \frac{a_i}{b_i}}{n-1}+\frac{a_n}{b_n})

于是

(∑i=1naibi)−(n∑i=1nai∑i=1nbi)≥∑i=1n−1aibi+anbn−n2(∑i=1n−1aibin−1+anbn)=n−22(n−1)(∑i=1n−1aibi)−n−22(anbn)≥n−22(n−1)(∑i=1n−1anbn)−n−22(anbn)=n−22(anbn)−n−22(anbn)=0(\sum_{i=1}^{n} \frac{a_i}{b_i})-(n\frac{\sum_{i=1}^{n}a_i}{\sum_{i=1}^{n}b_i}) \geq \sum_{i=1}^{n-1} \frac{a_i}{b_i}+\frac{a_n}{b_n} - \frac{n}{2}(\frac{\sum_{i=1}^{n-1} \frac{a_i}{b_i}}{n-1}+\frac{a_n}{b_n}) = \frac{n-2}{2(n-1)}(\sum_{i=1}^{n-1} \frac{a_i}{b_i}) - \frac{n-2}{2}(\frac{a_n}{b_n})\geq \frac{n-2}{2(n-1)}(\sum_{i=1}^{n-1} \frac{a_n}{b_n})-\frac{n-2}{2}(\frac{a_n}{b_n})=\frac{n-2}{2}(\frac{a_n}{b_n})-\frac{n-2}{2}(\frac{a_n}{b_n})=0

即

∑i=1naibi≥n∑i=1nai∑i=1nbi\sum_{i=1}^{n} \frac{a_i}{b_i} \geq n\frac{\sum_{i=1}^{n}a_i}{\sum_{i=1}^{n}b_i}

综上，原命题得证
①由二元形式（已证）得到
该定理十分有用
Exp1:Exp1:

∑cycab+c≥3×a+b+c2a+2b+2c=32\sum_{cyc}{}\frac{a}{b+c}\geq3\times\frac{a+b+c}{2a+2b+2c}=\frac{3}{2}

Exp2:Exp2:
已知abc=1abc=1 , a,b,c∈R+a,b,c\in R^+
则a+b+c≥3abc3=3a+b+c\geq 3\sqrt[3]{abc}=3
则∑cyca1+b+c≥1\sum_{cyc}{}\frac{a}{1+b+c}\geq 1
  ⟺  3×a+b+c3+2a+2b+2c≥1\iff 3\times\frac{a+b+c}{3+2a+2b+2c}\geq1
  ⟺  a+b+c≥3\iff a+b+c\geq3
注：使用前注意不失一般性，是否可以满足a1≥a2≥...≥ana_1\geq a_2 \geq ...\geq a_n , b1≤b2≤...≤bnb_1\leq b_2\leq...\le b_n (ai,bi∈R+)(a_i,b_i\in R^+)

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