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Virgil Nicula · Euler Joined: 28 Sep 2007 Posts: 618

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 $\underline {\overline {\left|\ bc\ \ge\ 0\ \Longrightarrow\ (b+c)^2(b^4+c^4)\ \ge\ (b^2+c^2)^3\ \right|}}$ .

Mateescu Constantin · Newton Joined: 21 Apr 2009 Posts: 306 Location: Pitesti

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 $b^2+bc+c^2\ge 0\ \Longrightarrow\ M\ge 0$ .
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Claudiu Mindrila · Fermat Joined: 01 Oct 2007 Posts: 517 Location: Targoviste

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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