Fie \( z_1,z_2\in\mathbb{C} \).Demonstrati urmatoarea inegalitate
\( |z_1|+|z_2|\le |z_1+z_2|+\frac{2|z_1z_2|}{|z_1+z_2|} \)
Inegalitate cu numere complexe
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DrAGos Calinescu
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Marius Mainea
- Gauss
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Prin ridicare la patrat
\( |z_1|^2+|z_2|^2+2|z_1z_2|\le|z_1+z_2|^2+4|z_1z_2|+\frac{4|z_1z_2|^2}{|z_1+z_2|^2} \)
sau
\( 0\le z_1\overline{z_2}+z_2\overline{z_1}+2|z_1z_2|+\frac{4|z_1z_2|^2}{|z_1+z_2|^2} \)
care este adevarata deoarece
\( -z_1\overline{z_2}-z_2\overline{z_1}\le|z_1\overline{z_2}+z_2\overline{z_1}|\le 2|z_1z_2| \)
\( |z_1|^2+|z_2|^2+2|z_1z_2|\le|z_1+z_2|^2+4|z_1z_2|+\frac{4|z_1z_2|^2}{|z_1+z_2|^2} \)
sau
\( 0\le z_1\overline{z_2}+z_2\overline{z_1}+2|z_1z_2|+\frac{4|z_1z_2|^2}{|z_1+z_2|^2} \)
care este adevarata deoarece
\( -z_1\overline{z_2}-z_2\overline{z_1}\le|z_1\overline{z_2}+z_2\overline{z_1}|\le 2|z_1z_2| \)
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andy crisan
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Deomnstram
\( (z_1\overline{z_2}+z_2\overline{z_1})(z_1\overline{z_2}+z_2\overline{z_1})=2|z_1|^2|z_2|^2+z_1^2\overline{z_2}^2+z_2^2\overline{z_1}^2\leq4|z_1|^2|z_2|^2\Leftrightarrow 2|z_1|^2|z_2|^2-z_1^2\overline{z_2}^2-z_2^2\overline{z_1}^2=|z_1\overline{z_2}-z_2\overline{z_1}|^2\geq0 \). inegalitate adevarata.
Ridicand la patrat obtinemMarius Mainea wrote: \( |z_1\overline{z_2}+z_2\overline{z_1}|\le 2|z_1z_2| \)
\( (z_1\overline{z_2}+z_2\overline{z_1})(z_1\overline{z_2}+z_2\overline{z_1})=2|z_1|^2|z_2|^2+z_1^2\overline{z_2}^2+z_2^2\overline{z_1}^2\leq4|z_1|^2|z_2|^2\Leftrightarrow 2|z_1|^2|z_2|^2-z_1^2\overline{z_2}^2-z_2^2\overline{z_1}^2=|z_1\overline{z_2}-z_2\overline{z_1}|^2\geq0 \). inegalitate adevarata.