MT [177] Three product sums

For any two full permutations of 1,2,3,4,5,6 $(a_1,a_2,a_3,a_4,a_5,a_6)$ and $(b_1,b_2,b_3,b_4,b_5,b_6)$, Find the minimum value of $\displaystyle S=\sum_{i=1}^6 ia_ib_i$______


Solution: $\displaystyle\sum_{i=1}^6 ia_ib_i \ge6\sqrt[6]{6!}=72\sqrt{5}>160.$
and $162=1*5*5+2*4* 4+3*3*3+4*6*1+5*1*6+6*2*2$
and let the maximum number in $ia_ib_i$ be $x(\textbf{easy to know}x\ge30)$, So $\displaystyle\sum_{i=1}^6 ia_ib_i\ge x+5\sqrt[5]{\frac{6!}{x}}\ge 30+60\sqrt[5]{50}>161 $ So the minimum value is 162

Guess you like

Origin http://43.154.161.224:23101/article/api/json?id=325198436&siteId=291194637