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Posted: **Mon Apr 06, 2009 9:29 pm**

Fie \( a_i > 0 \), \( i=\overline{1,n} \) astfel incat
\( \sum_{k=1}^n \frac{1}{a_k}=1 \)

Demonstrati ca \( \forall z_i \in \mathbb{C} \), \( i=\overline{1,n} \)
\( \sum_{k=1}^na_k det{z_k^2} \geq det{ \sum_{k=1}^n z_k}^2 \)

Posted: **Mon Apr 06, 2009 9:43 pm**

\( \sum_{k=1}^na_k det{z_k^2}=\sum_{k=1}^na_k det{z_k^2} \cdot \sum_{i=1}^n \frac{1}{a_i}\geq (\sum_{k=1}^n |z_k|)^2\geq | \sum_{k=1}^n z_k|^2 \)

Am aplicat un CBS si inegalitatea modulului.

Posted: **Mon Apr 06, 2009 9:52 pm**

\( |\sum_{k=1}^n z_k|^2\le (\sum_{k=1}^n |z_k|)^2=(\sum_{k=1}^n\frac{1}{\sqrt{a_k}}\sqrt{a_k}|z_k|)^2\le\sum_{k=1}^n (\frac{1}{\sqrt{a_k}})^2\cdot\sum_{k=1}^n (\sqrt{a_k}|z_k|)^2=\sum_{k=1}^n a_k|z_k|^2 \).

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Inegalitatea Archbold

Inegalitatea Archbold

Re: inegalitatea archbold