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: $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$
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in the title. $\endgroup$