Edited .......

By $AM - GM$ we have that $$a^3+2b^2+2b+4\geq \frac{3}{2}a^2+2b^2+2b+\frac{7}{2}=\frac{3}{2}(a^2+b^2)+\frac{1}{2}(b^2+1)+2b+3\geq \frac{3}{2}\cdot 2ab+\frac{1}{2}\cdot 2b+2b+3= 3(ab+b+1)$$Similarly and use $abc=1$ we have $$\dfrac{1}{a^3+2b^2+2b+4}+\dfrac{1}{b^3+2c^2+2c+4}+\dfrac{1}{c^3+2a^2+2a+4}\leq\frac{1}{3}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)=\frac{1}{3}\ (\text{as needed})$$

It should be $\sum_{cyc}\frac {1}{3a+6b} $ not $\sum_{cyc}\frac {1}{3a+6} $

Let $a,b$ and $c$ be positive real numbers. Prove that$$\dfrac{a}{a^3+2b^2+2b+4}+\dfrac{b}{b^3+2c^2+2c+4}+\dfrac{c}{c^3+2a^2+2a+4}\leq \dfrac13$$Let $a,b$ and $c$ be positive real numbers such that $a+b+c=3.$ prove that $$\dfrac{a^2}{a^3+2b^2+5b+4}+\dfrac{b^2}{b^3+2c^2+5c+4}+\dfrac{c^2}{c^3+2a^2+5a+4} \leq\frac{3}{4(ab+bc+ca)}$$

sqing wrote: Let $a,b$ and $c$ be positive real numbers. Prove that$$\dfrac{a}{a^3+2b^2+2b+4}+\dfrac{b}{b^3+2c^2+2c+4}+\dfrac{c}{c^3+2a^2+2a+4}\leq \dfrac13$$ Let $a,b$ and $c$ be positive real numbers. Prove that $$\dfrac{a}{b^3+2a^2+2a+4}+\dfrac{b}{c^3+2b^2+2b+4}+\dfrac{c}{a^3+2c^2+2c+4}\leq \dfrac13$$

sqing wrote: Let $a,b$ and $c$ be positive real numbers. Prove that $$\dfrac{a}{b^3+2a^2+2a+4}+\dfrac{b}{c^3+2b^2+2b+4}+\dfrac{c}{a^3+2c^2+2c+4}\leq \dfrac13$$ $$\dfrac{a}{b^3+2a^2+2a+4}+\dfrac{b}{c^3+2b^2+2b+4}+\dfrac{c}{a^3+2c^2+2c+4}\leq\dfrac13\left(\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\right)\leq \dfrac13$$$$\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\leq 1\iff \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\geq 3$$ Let a,b,c be positive real numbers such that $abc=1$.Prove that $$ \dfrac{1}{{2{a^2} + {b^2} + 3}} + \dfrac{1}{{2{b^2} + {c^2} + 3}} + \dfrac{1}{{2{c^2} + {a^2} + 3}} \leq \frac{1}{2}$$

sqing wrote: Let a,b,c be positive real number such that $abc=1$.Prove that $$ \dfrac{1}{{2{a^2} + {b^2} + 3}} + \dfrac{1}{{2{b^2} + {c^2} + 3}} + \dfrac{1}{{2{c^2} + {a^2} + 3}} \leq \frac{1}{2}$$ Note that $$\sum_{cyc}\frac{1}{2a^2+b^2+3}=\sum_{cyc}\frac{1}{(b^2+a^2)+(a^2+1)+2}\leq\sum_{cyc}\frac{1}{2(ab+a+1)}=\frac{1}{2}$$

By AM-GM, $$\sum_{cyc}\dfrac{1}{a^3+2b^2+2b+4}=2\sum_{cyc}\dfrac{1}{(a^3+a^3+1)+3b^2+(b^2+1)+4b+6}$$$$\leq \frac{2}{3}\sum_{cyc}\dfrac{1}{(a^2+b^2)+2b+2}\leq\frac{1}{3}\sum_{cyc}\frac{1}{ab+b+1}=\frac{1}{3}.$$