Olimpíadas ⇒ (Balcânica - 2001) Inequação

[matbatrobin]

Sejam a, b, c reais positivos tais que [tex3]a+b+c\geq abc[/tex3]. Prove que:

• [tex3]a^2+b^2+c^2\geq \sqrt{3} abc[/tex3]

[triplebig]

Por rearranjo:

[tex3]a^2+b^2+c^2\geq ab+bc+ac\stackrel{*}{\geq} \sqrt{3abc(a+b+c)}\geq abc\sqrt{3}[/tex3]

*

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[tex3](ab)^2+(bc)^2+(ac)^2\geq abc(a+b+c)\\
\therefore\;(ab)^2+(bc)^2+(ac)^2+2abc(a+b+c)\geq 3abc(a+b+c)\;\\
\therefore\;ab+bc+ac\geq \sqrt{3abc(a+b+c)}[/tex3]