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V

vodichhocmai

Cho a,b,c\geq0. a+b+c=1. Chứng minh:
1a2+b2+c2+1ab+1bc+1ca30\frac{1}{a^2+b^2+c^2} + \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\geq30

Chú ý rằng đẳng thức xảy ra khi :[TEX]\left{\frac{1}{a^2+b^2+c^2}=3\\ \frac{1}{ab} = \frac{1}{bc} = \frac{1}{ca}=9[/TEX]

[TEX]LHS\ge \frac{1}{a^2+b^2+c^2} +\frac{9}{ab+bc+ca}[/TEX]

[TEX]LHS\ge \frac{1}{1-2(ab+bc+ca)}+\frac{9}{ab+bc+ca}[/TEX]

[TEX]f(x)=\frac{1}{1-2x}+\frac{9}{x}\ge f13\frac{1}{3}\ \ \ \ 0<x\le \frac{1}{3}[/TEX]

[TEX]Done!![/TEX]
 
B

bigbang195

Chú ý rằng đẳng thức xảy ra khi :[TEX]\left{\frac{1}{a^2+b^2+c^2}=3\\ \frac{1}{ab} = \frac{1}{bc} = \frac{1}{ca}=9[/TEX]

[TEX]LHS\ge \frac{1}{a^2+b^2+c^2} +\frac{9}{ab+bc+ca}[/TEX]

[TEX]LHS\ge \frac{1}{1-2(ab+bc+ca)}+\frac{9}{ab+bc+ca}[/TEX]

[TEX]f(x)=\frac{1}{1-2x}+\frac{9}{x}\ge f13\frac{1}{3}\ \ \ \ 0<x\le \frac{1}{3}[/TEX]

[TEX]Done!![/TEX]

[TEX]\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{7}{ab+bc+ac} \ge \frac{9}{(a+b+c)^2}+\frac{7}{\frac{(a+b+c)^2}{3}} =30[/TEX]
 
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