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 :] 
 A: 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.
A: 

This is what I have although it is incorrect any help would be greatly appreciated.!


$\mathbf{1.)}$ Consider now the expression $\frac{a^3+b^3+c^3}{3}$
The expressions has minimum value given by:
$a\geq b \geq c > 0 \implies a^3 \geq b^3 \geq c^3$
Thus $\frac{a^3+b^3+c^3}{3}$ has minimum value $c^3$
$\mathbf{2.)}$ On the other hand, consider the expression $\frac{a^2+bc}{b+c}$
Again we have $a \geq b \geq c >0 \implies a^2+bc \geq c^2+c\cdot c = 2c^2$
$ a \geq b \geq c > 0 \implies a^2+bc \leq a^2 + a \cdot a = 2a^2$
$ a \geq b \geq c > 0 \implies b+c \geq 2c $
and $ a \geq b \geq c > 0 \implies b+c \leq 2a$
$\mathbf{3.)}$ Finally we know that a positive fraction is maximized when the numerator is as large as possible and the denominator is as small as possible 
Therefore we have $ \frac{a^2+bc}{b+c} \leq \frac{2a^2}{2c} = \frac{a^2}{c} \leq a $
Using identical considerations, we have $\frac{b^2+ca}{ca} \leq \frac{a^2}{c} \leq a  $ and $\frac{c^2+ab}{a+b}\leq \frac{a^2}{c} \leq a$
So that we have
Hence, $\frac{a^2+bc}{b+c} \cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \leq a^3$
This is the maximum value of the expression
$\mathbf{4.)}$ The minimum value will be achieved for a fraction when the numerator is as small as possible and denominator is as large as possible
Again, by an overall similar consideration, we have $\frac{a^2+bc}{b+c} \geq \frac{a^2+c^2}{2b}$ is the minimum value of this expression
Similarly $\frac{b^2+ca}{c+a} \geq \frac{b^2+c^2}{2a}$ and $\frac{c^2+ab}{a+b} \geq \frac{c^2+b^2}{2a}$
Hence 
But we have $a^2+b^2 \geq 2b^2$, $c^2+b^2 \geq 2c^2$
Hence, 
Therefore, $\frac{a^2+bc}{b+c} \cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \geq \frac{bc^4}{a^2}$
$\mathbf{5.)}$ And finally, we have $a^3 \geq abc$ is the maximum value of $abc$
The minimum value of $\frac{a^3+b^3+c^3}{3}$ is $c^3$ which is greater than $a^3$ which is the maximum value of $\frac{a^2+bc}{b+c}\cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b}$
That is,
So 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} $
Again we have $\frac{a^2+bc}{b+c}\cdot \frac{b^2+ca}{c+a} \cdot \frac{c^2+ab}{a+b} \geq \frac{bc^4}{a^2}$ 
A: The right inequality.
By AM-GM we obtain:
$$\prod\limits_{cyc}(a^2+bc)=\sqrt[3]{\frac{\prod\limits_{cyc}((a^2+bc)(b^2+ac))^2}{\prod\limits_{cyc}(a^2+bc)}}=\sqrt[3]{\frac{\prod\limits_{cyc}(ab(c^2+ab)+c(a^3+b^3))^2}{\prod\limits_{cyc}(a^2+bc)}}\geq$$
$$\geq\sqrt[3]{\frac{\prod\limits_{cyc}(4(a^3+b^3)(c^2+ab)abc)}{\prod\limits_{cyc}(a^2+bc)}}\geq\sqrt[3]{\frac{\prod\limits_{cyc}((a+b)^3(c^2+ab)abc)}{\prod\limits_{cyc}(a^2+bc)}}=abc\prod\limits_{cyc}(a+b)$$
