cho a, b, c dương thõa [tex]\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2[/tex] . Chứng minh: [tex]\frac{1}{4a+1}+\frac{1}{4b+1}+\frac{1}{4c+1}\geq 1[/tex]
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$
$\Leftrightarrow \frac{4}{4a+4}+\frac{4}{4b+4}+\frac{4}{4c+4}=2$
$\Leftrightarrow 2= \frac{4}{4a+1+3}+\frac{4}{4b+1+3}+\frac{4}{4c+1+3}\leq \sum \frac{1}{4a+1} +1$
$\Leftrightarrow \sum \frac{1}{4a+1} \geq 2-1=1 \Rightarrow \square .$