$\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b
Started By lilolilo, 17-01-2014 - 22:02
#1
Posted 17-01-2014 - 22:02
#2
Posted 17-01-2014 - 22:05
bài này chắc bạn thiếu điều kiện a,b,c là số dương
$\frac{a}{1+a^{2}}\leq \frac{a}{2a}=\frac{1}{2}$
suy ra $VT\leq \frac{3}{2}$
$VP\geq \frac{3}{2}$ (bdt nesebit)
đpcm
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