$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$ 
Let $a$, $b$, and $c$ be positive real numbers with $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that:
  $$
  ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc.
$$

Let us consider the following proofs. 
$$
  a^{2}+b^{2}+c^{2} \geq ab+bc+ca
$$
By the Arithmetic Mean-Geometric Mean Inequality we have 
$$
a^{2}+b^{2} \geq 2ab,\ \  b^{2}+c^{2} \geq 2bc,\ \ c^{2}+a^{2} \geq 2ca \tag{1}
$$
If we add together all the inequalities $(1)$, we obtain
$$
  2a^{2}+2b^{2}+2c^{2} \geq 2ab+2bc+2ca
$$
By dividing both side by $2$, the result follows.
Now let us consider,
 $$
  ab(a+b) + bc(b+c) + ac(a+c) \geq 6abc \tag{2}
$$
I already have proved $(2)$
Then, 
We are given, 
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3 \implies bc+ac+ab=3abc \tag{3}
$$
Notice that we have 
$$
a^{2}+b^{2}+c^{2} \geq bc+ac+ab=3abc 
$$
So, 
$$
a^{2}+b^{2}+c^{2} \geq 3abc \tag{4}
$$
Let us multiply both side of $(4)$ by $\displaystyle\frac{2}{3}$, yield
$$
\frac{2}{3}(a^{2}+b^{2}+c^{2}) \geq 2abc 
$$
Here where I stopped. Would someone help me out ! Thank you so much 
 A: We note that
\begin{align*}
ab(a+b)+bc(b+c)+ac(a+c) &= a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\\
&=\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+
\frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}\\
&\ge \frac{4(a+b+c)^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})} \qquad \mbox{(by the Schwarz inequality)}\\
&=\frac{2}{3}(a+b+c)^2\\
&=\frac{2}{3}(a^2+b^2+c^2+2ab+2bc+2ac)\\
&=\frac{2}{3}(a^2+b^2+c^2) + \frac{4}{3}(ab+bc+ca)\\
&=\frac{2}{3}(a^2+b^2+c^2) + 4abc.
\end{align*}
Here, we employed the Schwarz inequality of the from
\begin{align*}
&\ (a+b+b+c+a+c)^2 \\
=&\ \left(\frac{a}{\sqrt{\frac{1}{b}}}\sqrt{\frac{1}{b}}+\frac{b}{\sqrt{\frac{1}{a}}}\sqrt{\frac{1}{a}}+\frac{b}{\sqrt{\frac{1}{c}}}\sqrt{\frac{1}{c}}+\frac{c}{\sqrt{\frac{1}{b}}}\sqrt{\frac{1}{b}}+\frac{a}{\sqrt{\frac{1}{c}}}\sqrt{\frac{1}{c}}+\frac{c}{\sqrt{\frac{1}{a}}}\sqrt{\frac{1}{a}}\right)^2\\
\le&\ \left(\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+
\frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}\right)\left(2\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \big)\right).
\end{align*}
A: We need to prove that
$$\sum\limits_{cyc}(a^2b+a^2c)\geq\frac{2(a^2+b^2+c^2)}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+4abc$$ or
$$\sum\limits_{cyc}(a^2b+a^2c)\geq\frac{2abc(a^2+b^2+c^2)}{ab+ac+bc}+4abc$$ or
$$\sum\limits_{cyc}ab\sum\limits_{cyc}(a^2b+a^2c)\geq2abc(a^2+b^2+c^2)+4abc(ab+ac+bc)$$ or
$$\sum_{cyc}(a^3b^2+a^3c^2+2a^3bc+2a^2b^2c)\geq\sum\limits_{cyc}
(2a^3bc+4a^2b^2c)$$
$$\sum\limits_{cyc}(a^3b^2+a^3c^2-2a^2b^2c)\geq0$$ or
$$\sum\limits_{cyc}c^2(a+b)(a-b)^2\geq0$$
Done!
