ta có: $\frac{1}{a^{2}+2b^{2}+3}=\frac{1}{a^{2}+b^{2}+b^{2}+1+2}\leq\frac{1}{2}.\frac{1}{ab+b+1}$CMTT ta được:$\frac{1}{a^{2}+2b^{2}+3}+\frac{1}{b^{2}+2c^{2}+3}+\frac{1}{c^{2}+2a^{2}+3}\leq\frac{1}{2}.(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1})=\frac{1}{2}.\frac{ab+b+1}{ab+b+1}=\frac{1}{2}$
ta có: $\frac{1}{a^{2}+2b^{2}+3}=\frac{1}{a^{2}+b^{2}+b^{2}+1+2}\leq\frac{1}{2ab+2b+2}=\frac{1}{2}.\frac{1}{ab+b+1}$.CMTT ta được:$\frac{1}{a^{2}+2b^{2}+3}+\frac{1}{b^{2}+2c^{2}+3}+\frac{1}{c^{2}+2a^{2}+3}\leg\frac{1}{2}.(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1})=\frac{1}{2}$
ta có: $\frac{1}{a^{2}+2b^{2}+3}=\frac{1}{a^{2}+b^{2}+b^{2}+1+2}\leq\frac{1}{2}.\frac{1}{ab+b+1}$CMTT ta được:$\frac{1}{a^{2}+2b^{2}+3}+\frac{1}{b^{2}+2c^{2}+3}+\frac{1}{c^{2}+2a^{2}+3}\le
q\frac{1}{2}.(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1})
=\frac{1}{2}.\frac{ab+b+1}{ab+b+1}=\frac{1}{2}$