$5b)$ $P=\frac{1}{a^{2}+2b^{2}+3}+ \frac{1}{b^{2}+2c^{2}+3} + \frac{1}{c^{2}+2a^{2}+3} \Leftrightarrow 2P= \frac{2}{a^{2}+2b^{2}+3}+ \frac{2}{b^{2}+2c^{2}+3} + \frac{2}{c^{2}+2a^{2}+3}$
Chứng minh theo câu $5a)$ nên ta có $:$
$2P \leq \frac{1}{ab+b+1}+ \frac{1}{bc+c+1} + \frac{1}{ca+a+1}= \frac{abc}{ab+b+abc}+ \frac{a}{abc+ac+a} + \frac{1}{ca+a+1}$ $($do $abc=1$$)$
$\Leftrightarrow 2P \leq \frac{ac}{ca+c+1}+ \frac{a}{ca+a+1} + \frac{1}{ca+a+1}= \frac{ca+a+1}{ca+a+1}=1 \Leftrightarrow P \leq \frac{1}{2}$
Khi $a=b=c=1$ thì $P=\frac{1}{1^{2}+2.1^{2}+3}+ \frac{1}{1^{2}+2.1^{2}+3} + \frac{1}{1^{2}+2.1^{2}+3}= \frac{1}{2}$
Vậy $P_{Max}= \frac{1}{2}$ khi $a=b=c=1$$.$