When a=b=c

$LHS=\dfrac{a^2}{9abc+a}+\dfrac{b^2}{9abc+b}+\dfrac{c^2}{9abc+c} \geq \dfrac{(a+b+c)^2}{27abc+a+b+c} \geq \dfrac{(a+b+c)^2}{(a+b+c)^3+a+b+c}=RHS$ $Q.E.D$

Prove that for positive real numbers $a, b, c$ the following inequality holds: \begin{align*} \frac{a}{9bc+1}+\frac{b}{9ca+1}+\frac{c}{9ab+1}\geq \frac{a+b+c}{1+3(ab+bc+ca)} \\ \frac{a^2}{9bc+1}+\frac{b^2}{9ca+1}+\frac{c^2}{9ab+1}\geq \frac{a^2+b^2+c^2}{1+3(ab+bc+ca)} \end{align*}

For the first one above: $LHS=\dfrac{a^2}{9abc+a}+\dfrac{b^2}{9abc+b}+\dfrac{c^2}{9abc+c} \geq \dfrac{(a+b+c)^2}{27abc+a+b+c}$ So we just need to prove that $ \dfrac{(a+b+c)^2}{27abc+a+b+c} \geq \dfrac{a+b+c}{1+3(ab+bc+ca)} \implies a+b+c+3(ab+bc+ca)(a+b+c) \geq a+b+c +27abc \implies (ab+bc+ca)(a+b+c) \geq 9abc$ Which is just $AM-GM$ so we are done.

$$\text{My Solution:}$$$$\frac{a}{9bc+1}+\frac{b}{9ca+1}+\frac{c}{9ab+1}\geq \frac{a+b+c}{1+(a+b+c)^2}$$By C-S $$\frac{a^2}{9abc+a}+\frac{b^2}{9cba+b}+\frac{c^2}{9cab+c}\geq \frac{(a+b+c)^2}{27abc+a+b+c}$$$$\text{Let a+b+c=p}$$By AM-GM $$p^3 \geq 27abc$$So,we have $$\frac{p}{p^3+p} \geq \frac{1}{1+p^2}$$and they are equal The equality case occurs when a=b=c