có $a^{2}$ +$2b^{2}$ +3 = $a^{2}$ +$b^{2}$ +$b^{2}$ +1 +2 $\geq$ 2ab +2b+2
=> $\frac{1}{a^{2} +2b^{2} +3}$ $\leq$ $\frac{1}{2ab +2b+2}$
Tương tự nốt 2 cái kia .
=>P $\leq$ $\frac{1}{2}$$\left ( \frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+c+1} \right )$
Có abc=1 ,$\frac{1}{ab+b+1}$ + $\frac{1}{bc+c+1}$ + $\frac{1}{ca+c+1}$
= $\frac{1}{ab+b+1}$ + $\frac{abc}{bc+c+abc}$ $\frac{1}{\frac{1}{b}+a+1}$
= $\frac{1}{ab+b+1}$ +$\frac{ab}{ab+b+1}$ +$\frac{b}{abc +ab+1}$
=$\frac{1+ab+b}{ab+b+1}$=1
=> P$\leq$ $\frac{1}{2}$