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Posted: **Tue Mar 11, 2008 12:37 pm**

Se stie ca \( \left\|\ \begin{array}{ccc}
(b+c)^2 & \le & 2(b^2+c^2)\\\\
2(b^4+c^4) & \ge & (b^2+c^2)^2\end{array}\ \right\|\ \) . Sa se arate ca \( \underline {\overline {\left|\ bc\ \ge\ 0\ \Longrightarrow\ (b+c)^2(b^4+c^4)\ \ge\ (b^2+c^2)^3\ \right|}} \) .

Posted: **Tue Jan 26, 2010 4:25 pm**

Notam \( M=(b+c)^2(b^4+c^4)-(b^2+c^2)^3 \). Atunci \( M=(b^2+2bc+c^2)(b^4+c^4)-(b^6+3b^4c^2+3c^4b^2+c^6) \)

\( \Longleftrightarrow\ M=b^6+b^2c^4+2b^5c+2bc^5+c^2b^4+c^6-b^6-3b^4c^2-3c^4b^2-c^6=2b^5c+2bc^5-2b^2c^4-2b^4c^2 \)

\( \Longleftrightarrow\ M=2bc(b^4+c^4-b^3c-bc^3)=2bc\left\[(b^3(b-c)-c^3(b-c)\right\]\ \Longleftrightarrow\ \fbox{\ M=2bc(b-c)^2(b^2+bc+c^2)\ } \)

Cum \( bc\ge 0 \) , \( (b-c)^2\ge 0 \) si \( b^2+bc+c^2\ge 0\ \Longrightarrow\ M\ge 0 \).

Posted: **Thu Jan 28, 2010 9:52 pm**

Sau cu Holder:
\( \left(b^{4}+c^{4}\right)\left(b+c\right)\left(b+c\right)\ge\left(\sqrt[3]{b^{6}}+\sqrt[3]{c^{6}}\right)^{3}=\left(b^{2}+c^{2}\right)^{3} \)

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Comportarea "produsului" a doua inegalitati

Comportarea "produsului" a doua inegalitati