Câu 5. $ BDT \Leftrightarrow (\frac{1}{4-a} -\frac{1}{4} ) + (\frac{1}{4-b} -\frac{1}{4} ) + (\frac{1}{4-c } -\frac{1}{4} ) \geq \frac{a^2+b^2+c^2}{16} $
$ \Leftrightarrow \frac{a}{4(4-a)} + \frac{b}{4(4-b)} + \frac{c}{4(4-c)} \geq \frac{a^2+b^2+c^2}{16} $
Ta có: $ \frac{a}{4(4-a)} + \frac{a(4-a)}{16} \geq \frac{a}{4} $
Tương tự, ta có: $ \frac{a}{4(4-a)} + \frac{b}{4(4-b)} + \frac{c}{4(4-c)} \geq \frac{a}{4} + \frac{b}{4} + \frac{c}{4} - ( \frac{a(4-a)}{16} + \frac{b(4-b)}{16} + \frac{c(4-c)}{16} ) $
$ \Rightarrow \frac{a}{4(4-a)} + \frac{b}{4(4-b)} + \frac{c}{4(4-c)} \geq \frac{a}{4} + \frac{b}{4} + \frac{c}{4} - ( \frac{a}{4} + \frac{b}{4} + \frac{c}{4} -\frac{a^2+b^2+c^2}{16} ) $
$ \Rightarrow \frac{a}{4(4-a)} + \frac{b}{4(4-b)} + \frac{c}{4(4-c)} \geq \frac{a^2+b^2+c^2}{16} $
$\Rightarrow $ ĐPCM
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