Fie \( \sigma_k(x_1, ..., x_n) = \sum x_{i_1} \cdots x_{i_k} \) a \( k \)-a suma simetrica elementara.
Aratati ca pentru \( x, y, z, t > 0 \),
\( \sigma_2 \left( \sqrt{x}, \sqrt{y}, \sqrt{z}, \sqrt{t} \right) \ge 3\left( \sigma_3(x, y, z, t) \right)^{1/3} \).
[ DMO 2008, Problema 1 ]
Inegalitate simetrica non-standard
Moderators: Laurian Filip, Filip Chindea, Radu Titiu, maky, Cosmin Pohoata
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Filip Chindea
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Inegalitate simetrica non-standard
Life is complex: it has real and imaginary components.
Iata cea mai simpla si cea mai naturala solutie:
Din inegalitatea AM-GM se obtine ca:
\( {(\sqrt{xy}+\sqrt{zt})+(\sqrt{xz}+\sqrt{yt})+(\sqrt{xt}+\sqrt{yz})\geq 3\sqrt[3]{(\sqrt{xy}+\sqrt{zt})(\sqrt{xz}+\sqrt{yt})(\sqrt{xt}+\sqrt{yz})}}= \)
\( =3\sqrt[3]{xyz+yzt+ztx+txy+\sqrt{x^3yzt}+\sqrt{xy^3zt}+\sqrt{xyz^3t}+\sqrt{xyzt^3}}\geq \)
\( \geq 3\sqrt[3]{xyz+yzt+ztx+txy} \)
Din inegalitatea AM-GM se obtine ca:
\( {(\sqrt{xy}+\sqrt{zt})+(\sqrt{xz}+\sqrt{yt})+(\sqrt{xt}+\sqrt{yz})\geq 3\sqrt[3]{(\sqrt{xy}+\sqrt{zt})(\sqrt{xz}+\sqrt{yt})(\sqrt{xt}+\sqrt{yz})}}= \)
\( =3\sqrt[3]{xyz+yzt+ztx+txy+\sqrt{x^3yzt}+\sqrt{xy^3zt}+\sqrt{xyz^3t}+\sqrt{xyzt^3}}\geq \)
\( \geq 3\sqrt[3]{xyz+yzt+ztx+txy} \)