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O inegalitate frumoasa
Posted: Tue Mar 11, 2008 2:54 pm
by Marius Dragoi
Aratati ca pentru orice \( a,b,c>0 \) avem:
\( \sum_{ciclic}^{} \sqrt {a^4+a^2b^2+b^4} \geq \sum_{ciclic}^{} \sqrt {2a^4+a^2bc} \)
Gazeta Matematica
Posted: Wed Mar 12, 2008 11:00 pm
by Marius Dragoi
\( \sum_{ciclic}^{} {\sqrt {a^4+a^2b^2+b^4} = \sum_{ciclic}^{} {\sqrt {(a^4+\frac{a^2b^2}{2}) + (b^4+\frac{a^2b^2}{2})} \frac{Cauchy}{\geq} \)
\( {\frac{1}{\sqrt 2}{\sum_{ciclic}^{}{({\sqrt {a^4+\frac{a^2b^2}{2}}})+({\sqrt{b^4+\frac{a^2b^2}{2}}})} = {\frac{1}{\sqrt2}}{\sum_{ciclic}^{}{({\sqrt{a^4+\frac{a^2b^2}{2}}})+({\sqrt{a^4+\frac{a^2c^2}{2}}})} \frac{m_a-m_g}{\geq} \)
\( {\sqrt2}{\sum_{ciclic}^{}{({({a^4+ \frac{a^2b^2}{2}})({a^4+ \frac{a^2c^2}{2}})})}^{\frac{1}{4}} \frac{Cauchy}{\geq} {\sqrt 2}{\sum_{ciclic}^{}{\sqrt{a^4+\frac{a^2bc}{2}}} = \sum_{ciclic}^{}{\sqrt{2a^4+a^2bc}} \)