Cho a,b,c>0 , abc=1
CMR:
$\sum \frac{1}{a^2+2b^2+3}\leq \frac{1}{2}$
$\sum \frac{1}{a^2+2b^2+3}\leq \frac{1}{2}$
Started By VuQuyDat, 02-05-2018 - 21:44
#2
Posted 02-05-2018 - 22:13
Khoa Linh
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Thiếu úy
$\sum \frac{1}{a^2+2b^2+3}=\sum \frac{1}{(a^2+b^2)+(b^2+1)+2}\leq\frac{1}{2} \sum \frac{1}{ab+b+1}=\frac{1}{2}$
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