Let a, b, c be positive real numbers. Prove that Let a,b,c be positive real numbers. Prove that

$$\frac{a^3+b^3+c^3}{3}\geq\frac{a^2+bc}{b+c}\cdot\frac{b^2+ca}{c+a}\cdot\frac{c^2+ab}{a+b}\geq abc$$  

I will post what I had solved originally, however it is unfortunately incorrect. Please help solve and/or aid in finding my mistakes :] 

The left inequality.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since $\prod\limits_{cyc}(a^2+bc)=2a^2b^2c^2+\sum\limits_{cyc}(a^3b^3+a^4bc)=8w^6+A(u,v^2)w^3+B(u,v^2)$,
$\prod\limits_{cyc}(a+b)=9uv^2-w^3$ and $a^3+b^3+c^3=27u^3-27uv^2+3w^3$, we see that
our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function.
But  the concave function gets a minimal value for an extremal value of $w^3$,
which happens for equality case of two variables and we must check  $w^3\rightarrow0^+$.


*

*Let $w^3\rightarrow0^+$. Let $c\rightarrow0^+$.


We get $(a^3+b^3)(a+b)ab\geq3a^3b^3$, which is obvious;


*$b=c=1$, which gives $(a^2-1)^2(2a+1)\geq0$.


Done!
Also we can use a full expanding and we'll get something obvious:
$\sum\limits_{sym}\left(a^5b+a^4b^2-\frac{1}{2}a^4bc-\frac{3}{2}a^3b^3+a^3b^2c-a^2b^2c^2\right)\geq0$, which is Muirhead.
