Problem 76 (difficulty: 10/10)

Prove that for any sequence \(\displaystyle a_1,a_2,\ldots,a_n\) of positive real numbers,

\(\displaystyle \frac1{\frac1{a_1}} + \frac2{\frac1{a_1}+\frac1{a_2}} + \frac3{\frac1{a_1}+\frac1{a_2}+\frac1{a_3}} + \ldots + \frac{n}{\frac1{a_1}+\frac1{a_2}+\ldots+\frac1{a_n}} < 2 (a_1+a_2+\ldots+a_n). \)

(KöMaL N. 189., November 1998)

Solution:

Applying the weighted AM-HM inequality,

\(\displaystyle \sum_{k=1}^n \frac{k}{\frac1{a_1}+\frac1{a_2}+\ldots+\frac1{a_k}} = \sum_{k=1}^n \frac2{k+1} \cdot \frac{1+2+\ldots+k}{\frac1{a_1}+\frac2{2a_2}+\ldots+\frac{k}{ka_k}} \le \sum_{k=1}^n \frac2{k+1} \cdot \frac{1\cdot a_1+2\cdot 2a_2+\ldots+k\cdot ka_k}{1+2+\ldots+k} = \)

\(\displaystyle = \sum_{k=1}^n \frac4{k(k+1)^2} \sum_{i=1}^k i^2a_i = \sum_{i=1}^n i^2a_i \sum_{i=k}^n \frac4{k(k+1)^2} < \sum_{i=1}^n i^2a_i \sum_{i=k}^n \frac{2(2k+1)}{k^2(k+1)^2} = \)

\(\displaystyle = \sum_{i=1}^n i^2a_i \sum_{i=k}^n \left(\frac2{k^2}-\frac2{(k+1)^2}\right) < \sum_{i=1}^n i^2a_i \left(\frac2{i^2}-\frac2{(n+1)^2}\right) < \sum_{i=1}^n i^2a_i \cdot \frac2{i^2} = 2 \sum_{i=1}^n a_i. \)

Remark. The constant \(\displaystyle 2\) on the right-hand side is sharp. If \(\displaystyle a_i=\frac1i\) and \(\displaystyle n\) is sufficiently large, the ratio between the two sides can be arbitrarily close to \(\displaystyle 1\).

Give me another random problem!