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Inequação

Enviado: Sex 20 Fev, 2009 01:49
por triplebig
Sejam [tex3]a,b,c>0[/tex3] com [tex3]a+b+c=1[/tex3] . Mostre que

[tex3]\frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\geq2[/tex3]

Re: Inequação

Enviado: Sex 20 Fev, 2009 11:06
por Beastie
Por [tex3]\text{MA}\geq\,\text{MG}[/tex3] , [tex3]\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{a+c}{a+b}\geq\,3\sqrt[3]{\frac{a+b}{b+c}\cdot\,\frac{b+c}{c+a}\cdot\,\frac{a+c}{a+b}}=3[/tex3] [tex3]\text{(*)}[/tex3] .

Desenvolvendo [tex3]\text{(*)}[/tex3] , sabendo que [tex3]a+b+c=1[/tex3] :

[tex3]\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{a+c}{a+b}=\frac{1-c}{b+c}+\frac{1-a}{c+a}+\frac{1-b}{a +b}\geq\,3[/tex3]

[tex3]\Rightarrow\,-1+\frac{1-c}{b+c}+\frac{1-a}{c+a}+\frac{1-b}{a +b}\geq\,3-1=2[/tex3]

[tex3]\Rightarrow\,(-a-b-c)+\frac{1-c}{b+c}+\frac{1-a}{c+a}+\frac{1-b}{a +b}\geq\,2[/tex3]

[tex3]\Rightarrow\,(-a+\frac{1-c}{b+c})+(-b+\frac{1-a}{c+a})+(-c+\frac{1-b}{a +b})\geq\,2[/tex3]

[tex3]\Rightarrow\,\frac{-a(b+c)+1-c}{b+c}+\frac{-b(c+a)+1-a}{c+a}+\frac{-c(a+b)+1-b}{a+b}\geq\,2[/tex3]

[tex3]\Rightarrow\,\frac{-a(1-a)+1-c}{b+c}+\frac{-b(1-b)+1-a}{c+a}+\frac{-c(1-c)+1-b}{a+b}\geq\,2[/tex3]

[tex3]\Rightarrow\,\frac{a^2+(1-a-c)}{b+c}+\frac{b^2+(1-a-b)}{c+a}+\frac{c^2+(1-c-b)}{a+b}\geq\,2[/tex3]

[tex3]\Rightarrow\,\boxed{\frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\geq\,2}[/tex3]