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L

linhhuyenvuong

Có:
[TEX]\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2} \geq a-\frac{ab}{2}[/TEX]

tt \Rightarrow[TEX]VT \geq a+b+c-\frac{ab+bc+ac}{2}=3-\frac{ab+bc+ac}{2}[/TEX]

[TEX](a+b+c)^2 \geq3(ab+bc+ac)[/TEX]

\Rightarrow[TEX]VT \geq3-\frac{(a+b+c)^2}{6}=\frac{3}{2}[/TEX]
 
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