Inegalitate exponentiala non-standard

[Filip Chindea]

Fie \( n \) un intreg pozitiv si \( x, y > 0 \) reale cu \( x^n + y^n = 1 \). Aratati ca

\( \left( \sum_{k=1}^n \frac{1 + x^{2k}}{1 + x^{4k}} \right) \left( \sum_{k=1}^n \frac{1 + y^{2k}}{1 + y^{4k}} \right) < \frac{1}{(1-x)(1-y)} \) .

[ IMO Shortlist 2007, A3 ]

[Radu Titiu]

\( y^n=1-x^n=(1-x)\left( \sum_{k=1}^n x^{k-1} \right) \) de unde rezulta:

\( \frac{1}{1-x}=\frac{\sum_{k=1}^n x^{k-1}}{y^n}. \)

Analog

\( \frac{1}{1-y}=\frac{\sum_{k=1}^n y^{k-1}}{x^n}. \)

Inegalitatea de demonstrat devine:

\( \left( \sum_{k=1}^n \frac{1 + x^{2k}}{1 + x^{4k}} \right) \left( \sum_{k=1}^n \frac{1 + y^{2k}}{1 + y^{4k}} \right) <\frac{\sum_{k=1}^n x^{k-1}}{y^n} \cdot \frac{\sum_{k=1}^n y^{k-1}}{x^n} \)

echivalent cu:

\( \left( \sum_{k=1}^n \frac{1 + x^{2k}}{1 + x^{4k}} \right) \left( \sum_{k=1}^n \frac{1 + y^{2k}}{1 + y^{4k}} \right) < \left( \sum _{k=1}^n \frac{1}{x^k}\right)\left( \sum _{k=1}^n \frac{1}{y^k}\right) \).

In continuare voi arata ca \( \frac{1+x^{2k}}{1+x^{4k}}<\frac{1}{x^k} \) echivalenta cu \( x^k+x^{3k}<x^{4k}+1 \Leftrightarrow \) \( (x^k-1)(x^{3k}-1)>0 \). \( \qed \)