Cho a,b,c>0 , abc=1
CMR:
$\sum \frac{1}{a^2+2b^2+3}\leq \frac{1}{2}$
$\sum \frac{1}{a^2+2b^2+3}=\sum \frac{1}{(a^2+b^2)+(b^2+1)+2}\leq\frac{1}{2} \sum \frac{1}{ab+b+1}=\frac{1}{2}$
$\sqrt[LOVE]{MATH}$
"If I feel unhappy, I do mathematics to become happy. If I am happy, I
do mathematics to keep happy" - Alfréd Rényi
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