Ta có: $a^2+2b+1=a^2+1+2b+2\geq 2a+2b+2$$\Rightarrow P\leq \frac{1}{2}(\Sigma (\frac{a}{a+b+1})$
$\Rightarrow \frac{3}{2}-P\geq$ $\frac{1}{2}$$(\Sigma \frac{b+1}{a+b+1})$
A/d Cauchy-Schwarz ta được:
$Q=\Sigma \frac{b+1}{a+b+1}=\Sigma \frac{(b+1)^2}{(b+1)(a+b+1)}\geq \frac{(a+b+c+3)^2}{\Sigma (a+1)(a+c+1)}$
Ta có:
$\Sigma (a+1)(a+c+1)=(a^2+b^2+c^2+ab+bc+ca+3(a+b+c)+3=\frac{1}{2}(a^2+b^2+c^2)+ab+bc+ca+3(a+b+c)+\frac{9}{2}=\frac{1}{2}(a+b+c+3)^2$
Suy ra:
$Q\geq \frac{(a+b+c+3)^2}{\frac{1}{2}(a+b+c+3)^2}\Rightarrow .............$
$\Rightarrow P\leq \frac{1}{2}$