Prove the inequality $a^{2}+b^{2}+c^{2}+abc(a+b+c) \geq 2(ab+bc+ca)$, given $a^{2}b+b^{2}c+c^{2}a+a^{2}b^{2}c^{2}=4$ 
Let $a, b$ and $c$ be positive real numbers such that $a^{2}b+b^{2}c+c^{2}a+a^{2}b^{2}c^{2}=4$. Prove that
  \begin{equation}
a^{2}+b^{2}+c^{2}+abc(a+b+c) \geq 2(ab+bc+ca)
\end{equation} 

Let us consider that 
\begin{equation*}
\begin{split}
\text{$(a+b+c)^{2}$} &=\text{$(a+b+c)(a+b+c)$} \\
&=\text{$a^{2}+ab+ac+ba+b^{2}+bc+ca+cb+c^{2}$} \\
&=\text{$a^{2}+b^{2}+c^{2}+2ac+2ab+2bc$} \\
&=\text{$a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$}
\end{split}
\end{equation*} 
and consider 
\begin{equation*}
\begin{split}
\text{$a^{2}+b^{2}+c^{2}$} &=\text{$(a+b+c)^{2}-2(ab+bc+ca)$} 
\end{split}
\end{equation*} 
The L.H.S of the inequality (1) 
\begin{equation}
\begin{split}
\text{$a^{2}+b^{2}+c^{2}+abc(a+b+c) $}   
&=\text{$(a+b+c)^{2}-2(ab+bc+ca)+abc(a+b+c)$} \\
&=\text{$(a+b+c)^{2}+(a+b+c)abc-2(ab+bc+ca)$} \\
&=\text{$(a+b+c)\bigg[(a+b+c)+abc \bigg]-2(ab+bc+ca)$} \\
\end{split}
\end{equation}
I have not real proof anything yet. I have through out some work, but I have not think how would I use $a^{2}b+b^{2}c+c^{2}a+a^{2}b^{2}c^{2}=4$. Please If you any idea that would help, please write it down. 

We'll prove that $a+b+c\geq3$.
Let $a+b+c<3$, $a=kx$, $b=ky$ and $c=kz$, where $k>0$ and $x+y+z=3$.
Hence, $k<1$ and $4=a^2b+b^2c+c^2a+a^2b^2c^2=k^3(x^2y+y^2z+z^2x+k^3x^2y^2z^2)<x^2y+y^2z+z^2x+x^2y^2z^2$,
which is contradiction because we'll prove now that $x^2y+y^2z+z^2x+x^2y^2z^2\leq4$.
Indeed, we need to prove that $\frac{(x^2y+y^2z+z^2x)(x+y+z)^3}{27}+x^2y^2z^2\leq\frac{4(x+y+z)^6}{729}$.
But by Rearrangement $x^2y+y^2z+z^2x\leq\frac{4(x+y+z)^3}{27}-xyz$.
Hence, it remains to prove that $(x+y+z)^3\geq27xyz$, which is AM-GM.
Id est, $a+b+c\geq3$.
Hence, it's enough to prove that $9abc\geq(a+b+c)\sum\limits_{cyc}(2ab-a^2)$ or
$\sum\limits_{cyc}(a^3-a^2b-a^2c+abc)\geq0$, which is Schur.
Done!
