<s>[INST] 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:
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$

 [/INST] 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:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{24\sqrt[3]{abc}}{a+b+c}\geq11.$$
</s>