math_ Derivation of ordering inequality

Ranking inequalities

• There are two ordered sequences ( It can be understood as monotonous ( discrete ) function ), namely
  $$\begin{gathered} \{a_n\}and\{b_n\},Respectively\; satisfy: \\ {a}_i<a_{i+1} \\ {b}_i<b_{i+1} \\ i=1,2,3,\cdots ,n-1 \end{gathered}$$

• Proof conclusion

  • $$\begin{gathered} \sum _{i=1}^n{a}_i{b}_i\ge \sum _{i=1}^n{a}_{i}R(\{b_n\}{)}_i \\ among,R(\{b_n\})Represents\; the\; pairwise\; sequence\{b_n\}A\; disorder\; of(arbitrarily)array(ShuffleRandomize/shuffle) \\ And\; with\; subscriptiThe\; expression\; ofR\{b_n{\}}_{i}Express{b}_{n}An\; arbitrary\; disordered\; sequence\; of{c}_{n}Of\; theiElements \\ The\; writing\; here\; is\; a\; bit\; like\; function\; call\; in\; programming \\ (Set\; functionRReturns\; a\; sequence\; that\; is\; out\; of\; order\{c_n\}) \\ therefore,For\; the\; convenience\; of\; writing,We\; have\; an\; equivalent\; expression: \\ \sum _{i=1}^n{a}_i{b}_i\ge \sum _{i=1}^n{a}_i{c}_i \end{gathered}$$

    $$\begin{gathered} Record\; the\; sequence\{b_n\}BeforeiItems\; and\; are{s}_i=\sum _{k=1}^{k=i}{b}_k \\ Record\; the\; sequence\{c_n\}BeforeiItems\; and\; ares{r}_i=\sum _{k=1}^{k=i}{c}_k \\ \begin{array}{rl}& b_i=s_i-s_{i-1}(important);i\in N^{+},i=1,2,3,\cdots ,n\\ & s_i\le s{r}_{i}The\; expression\; of\; this\; inequality\; is,The\; sequence\{b_n\}ArbitrarilyiXiang\; He,The\; minimum\; is{s}_i\\ & That\; makes\; sense,After\; all{s}_{i}The\; value\; of\; is\; the\; smallest\; in\; the\; sequenceiThe\; sum\; of\; the\; elementss{r}_{i}No\; comparison{s}_{i}Small\\ & s_n=s{r}_{n}It\text{'}s\; easy\; to\; understand,The\; synthesis\; of\; a\; group\; of\; numbers\; out\; of\; order\; is\; always\; consistent(At\; this\; time,\; the\; partial\; sum\; is\; the\; total\; sum)\end{array} \end{gathered}$$

  • $$\begin{gathered} also:*\begin{array}{rl}& s_0=0\\ & s_1=b1\end{array} \\ for\; example:b_1=s_1-s_0=s_1 \\ {b}_2=s_2-s_1 \\ {b}_3=s_3-s_2 \\ Besides,because\{a_n\}The\; sequence\; is\; monotonous,It\text{'}s\; easy\; to\; know{a}_i-a_{i-1}\le 0 \\ Thus\; there\; are: \end{gathered}$$

    $$s_i(a_i-a_{i-1})\ge s{r}_i(a_i-a_{i-1})\text{\hspace{1em}(core)}$$

• One side
  $$\begin{gathered} \sum _{i=1}^n{a}_i{b}_i=\sum _{i=1}^n{a}_i(s_i-s_{i-1}) \\ Summation\; can\; be\; used\; to\; express\; addition\; concisely,But\; sometimes\; it\; is\; not\; conducive\; to\; see\; the\; law \\ We\; willi=1,2,3,\cdots ,nBring\; them\; to\; the\; right\; side\; of\; the\; equation,Observe\; the\; law \\ \begin{array}{rlr}{a}_1{b}_1& =a_1(s_1-s_0)& =a_1{s}_1-a_1{s}_0\\ a_2{b}_2& =a_2(s_2-s_1)& =a_2{s}_2-a_2{s}_1\\ a_3{b}_3& =a_3(s_3-s_2)& =a_3{s}_3-a_3{s}_2\\ ⋮& ⋮& ⋮\\ a_{n-1}{b}_{n-1}& =a_{n-1}(s_{n-1}-s_{n-2})& =a_{n-1}{s}_{n-1}-a_{n-1}{s}_{n-2}\\ a_n{b}_n& =a_n(s_n-s_{n-1})& =a_n{s}_n-a_n{s}_{n-1}\end{array} \end{gathered}$$
  
  $$\begin{gathered} Add\; the\; above\; equations,Or\; get \\ \sum _{i=1}^n{a}_i{b}_i=s_1(a_1-a_2)+s_2(a_2-a_3)+s_3(a_3-a_4)+\cdots \\ +s_{n-1}(a_{n-1}-a_n)+a_n{s}_n \\ =\left(\sum _{i=1}^{n-1}{s}_i(a_i-a_{i+1})\right)+a_n{s}_n \end{gathered}$$

• On the other hand , Similar to the above derivation , Yes Unordered product and

  • $$\sum _{i=1}^n{a}_i{c}_i=\left(\sum _{i=1}^{n-1}s{r}_i(a_i-a_{i+1})\right)+a_{n}s{r}_n$$

• So there is a conclusion

$$\begin{gathered} \because s_i(a_i-a_{i-1})\ge s{r}_i(a_i-a_{i-1}) \\ {s}_n=s{r}_n \\ \therefore \left(\sum _{i=1}^{n-1}{s}_i(a_i-a_{i+1})\right)+a_n{s}_n\ge \left(\sum _{i=1}^{n-1}s{r}_i(a_i-a_{i+1})\right)+a_{n}s{r}_n \\ \therefore \sum _{i=1}^n{a}_i{b}_i\ge \sum _{i=1}^n{a}_i{c}_i\text{\hspace{1em}(core)} \end{gathered}$$

The other half of the equation

• A similar principle , Can be derived Inverse product sum Less than Unordered product and Of

• $$\sum _{i=1}^n{a}_i{b}_i\ge \sum _{i=1}^n{a}_i{c}_i\ge \sum _{i=1}^n{a}_i{b}_{n-i+1}$$

summary