Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly.

Exercise 1.3.4(a) Let $a,b,c$ be positive real numbers. Prove that
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$
(b) For real numbers $a,b,c \gt0$ and $n \leq3$ prove that:
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$

 A: By C-S
$$\sum_{cyc}\frac{a}{b}=\sum_{cyc}\frac{a^2}{ab}\geq\frac{(a+b+c)^2}{ab+ac+bc}.$$
Thus, it's enough to prove that
$$\frac{(a+b+c)^2}{ab+ac+bc}+\frac{3\sqrt[3]{abc}}{a+b+c}\geq4.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(v^2)\geq0,$ where $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $v^2$, which happens for equality case of two variables.
Since the last inequality is homogeneous, we can assume $b=c=1$. Also, let $a=x^3$.
Id est, we need to prove that
$$\frac{(x^3+2)^2}{2x^3+1}+\frac{3x}{x^3+2}\geq4$$ or
$$(x-1)^2(x^6+2x^5+3x^4+2x^3+x^2+6x+3)\geq0.$$ 
Done!
The following inequality is also true.
Let $a$, $b$ and $c$ be positives. Prove that:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{24\sqrt[3]{abc}}{a+b+c}\geq11.$$
A: Write $$\frac ab+\frac ab+\frac bc\geq \frac{3a}{\sqrt[3]{abc}}$$ by AM-GM.
You get $$\operatorname{LHS} \geq \frac{a+b+c}{\sqrt[3]{abc}}+n\left(\frac{\sqrt[3]{abc}}{a+b+c}\right).$$
Set $$z:=\frac{a+b+c}{\sqrt[3]{abc}}$$ and then notice that for $n\leq 3$, $$z+\frac{3n}{z}\geq 3+n.$$  Indeed the minimum is reached for $z=\sqrt{3n}\leq 3$; since $z\geq 3$, the minimum is reached in fact for $z=3$.
A: Following above motivations and applying AM-GM three times:
\begin{align}
&\frac13\left(\frac ab+\frac ab+\frac bc\right)+\frac13\left(\frac bc+\frac bc+\frac ca\right)+\frac13\left(\frac ca+\frac ca+\frac ab\right)+\frac{3\sqrt[3]{abc}}{a+b+c}\\
&\ge \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{abc}}+\frac{3\sqrt[3]{abc}}{a+b+c}\\
&=\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{3\sqrt[3]{abc}}{a+b+c}\\
&\ge4\left(\left(\frac{a+b+c}{3\sqrt[3]{abc}}\right)^2\right)^\frac14\\
&\ge4.
\end{align}
