设 0<a1≤a2≤...≤an0<a_1\le a_2\le...\le a_n0<a1≤a2≤...≤an,0<b1≤b2≤...≤bn0<b_1\le b_2\le...\le b_n0<b1≤b2≤...≤bn,求证 a1bn+a2bn−1+...+anb1≥n⋅a1+a2+...+anb1+b2+...+bn\displaystyle\frac{a_1}{b_n}+\frac{a_2}{b_{n-1}}+...+\frac{a_n}{b_1}\ge n\cdot\frac{a_1+a_2+...+a_n}{b_1+b_2+...+b_n}bna1+bn−1a2+...+b1an≥n⋅b1+b2+...+bna1+a2+...+an