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S

soccan

$a^2+b^2 \ge 2ab\\
b^2+1 \ge 2b\\
\longrightarrow a^2+2b^2+3 \ge 2(ab+b+1)\\
\longrightarrow \sum \dfrac{1}{a^2+2b+3}\le\sum \dfrac{1}{2(ab+b+1)}=\dfrac{1}{2}(\dfrac{1}{ab+b+1}+\dfrac{ab}{babc+abc+ab}+\dfrac{b}{abc+ab+b})= \dfrac{1}{2}
$
 
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