Desigualdade entre as médias

por Luan, o Rocha Seg 20 maio - 23:58

Elementos da Matemática, Vol 0.
Capítulo 03, Médias.

45) Sendo a, b,c reais positivos tais que [latex]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1[/latex], mostre que [latex](a-1)(b-1)(c-1)\geq 8[/latex].

Gostaria de compartilhar a resolução com vocês:

[latex](a-1)(b-1)(c-1)\geq 8\Leftrightarrow (ab-a-b+1)(c-1)\geq 8\Leftrightarrow abc-ac-bc+c-ab+a+b-1\geq 8\Leftrightarrow abc+a+b+c-(ab+bc+ca+1)\geq 8\Leftrightarrow abc+a+b+c\geq ab+bc+ca+9\Leftrightarrow (abc+a+b+c)\frac{1}{abc}\geq (ab+bc+ca+9)\frac{1}{abc}\Leftrightarrow 1+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )+\frac{9}{abc}\Leftrightarrow 1+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 1+\frac{9}{abc}\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq \frac{9}{abc}\Leftrightarrow abc\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \right )\geq \frac{9}{abc}.abc\Leftrightarrow a+b+c\geq \frac{9}{1}\Leftrightarrow \frac{a+b+c}{3}\geq \frac{3}{1}\Leftrightarrow \frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\therefore M.A.\geq M.H.[/latex]

por **Giovana Martins** Dom 26 maio - 8:53

\[\mathrm{(a-1)(b-1)(c-1)\geq 8\leftrightarrow \left ( \frac{a-1}{a} \right )\left ( \frac{b-1}{b} \right )\left ( \frac{c-1}{c} \right )\geq \frac{8}{abc}}\]

\[\mathrm{Do\ enunciado\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\leftrightarrow 1-\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\ \therefore\ 1-\frac{1}{a}=\underset{M_A\ \geq \ M_G}{\underbrace{\mathrm{\frac{1}{b}+\frac{1}{c}\geq \frac{2}{\sqrt{bc}}}}}\ (i)}\]

\[\mathrm{Analogamente:1-\frac{1}{b}\geq \frac{2}{\sqrt{ac}}\ (ii)\ e\ 1-\frac{1}{c}\geq \frac{2}{\sqrt{ab}}\ (iii)}\]

\[\mathrm{Do\ produto\ entre\ (i),(ii)\ e\ (iii):\left ( 1-\frac{1}{a} \right )\left ( 1-\frac{1}{b} \right )\left ( 1-\frac{1}{c} \right )\geq \frac{8}{\sqrt{a^2b^2c^c}}}\]

\[\mathrm{\therefore \ \boxed{\mathrm{(a-1)(b-1)(c-1)\geq 8}}}\]

por Luan, o Rocha Dom 26 maio - 9:31

Giovana Martins escreveu:
\[\mathrm{(a-1)(b-1)(c-1)\geq 8\leftrightarrow \left ( \frac{a-1}{a} \right )\left ( \frac{b-1}{b} \right )\left ( \frac{c-1}{c} \right )\geq \frac{8}{abc}}\]

\[\mathrm{Do\ enunciado\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\leftrightarrow 1-\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\ \therefore\ 1-\frac{1}{a}=\underset{M_A\ \geq \ M_G}{\underbrace{\mathrm{\frac{1}{b}+\frac{1}{c}\geq \frac{2}{\sqrt{bc}}}}}\ (i)}\]

\[\mathrm{Analogamente:1-\frac{1}{b}\geq \frac{2}{\sqrt{ac}}\ (ii)\ e\ 1-\frac{1}{c}\geq \frac{2}{\sqrt{ab}}\ (iii)}\]

\[\mathrm{Do\ produto\ entre\ (i),(ii)\ e\ (iii):\left ( 1-\frac{1}{a} \right )\left ( 1-\frac{1}{b} \right )\left ( 1-\frac{1}{c} \right )\geq \frac{8}{\sqrt{a^2b^2c^c}}}\]

\[\mathrm{\therefore \ \boxed{\mathrm{(a-1)(b-1)(c-1)\geq 8}}}\]

Muito bom! Resultado mais simples e elegante Desigualdade entre as médias 1f44f