$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$

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*}

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!