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Problem A. 408. (October 2006)

A. 408. The positive reals a_1,a_2,\ldots,a_n and b1\leb2\le...\lebn satisfy a1+a2+...+ak\leb1+b2+...+bk for all 1\lek\len. Prove that

\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n} \ge \frac1{b_1}+\frac1{b_2}+\cdots+\frac1{b_n}.

(5 pont)

Deadline expired on November 15, 2006.

Solution. Applying the weighted AM-HM inequality,

\frac{\frac1{b_1}\cdot\frac{b_1}{a_1}+\dots+ \frac1{b_n}\cdot\frac{b_n}{a_n}}{\frac1{b_1}+\dots+\frac1{b_n}}\ge \frac{\frac1{b_1}+\dots+\frac1{b_n}}{ \frac1{b_1}\cdot\frac{a_1}{b_1}+\dots+ \frac1{b_n}\cdot\frac{a_n}{b_n}},

so

\frac1{a_1}+\dots+\frac1{a_n}\ge \frac{\left(\frac1{b_1}+\dots+\frac1{b_n}\right)^2}{ \frac{a_1}{b_1^2}+\dots+\frac{a_n}{b_n^2}}.

Estimate the last denominator as

\frac{a_1}{b_1^2}+\dots+\frac{a_n}{b_n^2}= \sum_{k=1}^{n-1}\left(\frac1{b_k^2}-\frac1{b_{k+1}^2}\right) (a_1+\dots+a_k)+\frac1{b_n^2}(a_1+\dots+a_n)\le

\le \sum_{k=1}^{n-1}\left(\frac1{b_k^2}-\frac1{b_{k+1}^2}\right) (b_1+\dots+b_k)+\frac1{b_n^2}(b_1+\dots+b_n)= \frac{b_1}{b_1^2}+\dots+\frac{b_n}{b_n^2}= \frac1{b_1}+\dots+\frac1{b_n}.

Therefore,

\frac1{a_1}+\dots+\frac1{a_n}\ge \frac{\left(\frac1{b_1}+\dots+\frac1{b_n}\right)^2}{ \frac{a_1}{b_1^2}+\dots+\frac{a_n}{b_n^2}} \ge \frac{\left(\frac1{b_1}+\dots+\frac1{b_n}\right)^2}{ \frac1{b_1}+\dots+\frac1{b_n}} = \frac1{b_1}+\dots+\frac1{b_n}.

Statistics:

14 students sent a solution.
5 points:Gyenizse Gergő, Hujter Bálint, Kisfaludi-Bak Sándor, Kónya 495 Gábor, Lovász László Miklós, Nagy 224 Csaba, Nagy 314 Dániel, Sümegi Károly, Tomon István, Wolosz János.
4 points:Korándi Dániel, Kornis Kristóf, Nagy 235 János.
3 points:1 student.

Problems in Mathematics of KöMaL, October 2006