Strange Cube Root Offense in an Inequality

Question

I don't know how to tackle the unusual cube root present in this inequality-
$1.$For real numbers $a,b,c > 0$ and $n\le3$ prove that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c}\right)\ge 3+n$$ Here is another question with the same lesser side (and of course I couldn't prove)-
$2.$Let $a, b, c$ be positive real numbers such that $a + b + c = ab + bc + ca$ and $n ≤ 3$. Prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\ge 3+n$$ What I attempted was this- $$\left(a+b+c\right)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\right)\ge \left(a+b+c\right)\left(3+n\right)$$ Avoiding the RHS for some time-
$$\left(a+b+c\right)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2}\right)\ge (a+b+c)^2+\frac{3n(a+b+c)}{a^2+b^2+c^2}$$ After this step I don't know where to use $a+b+c=ab+bc+ca$.
These are very basic. I need a solution using AM-GM Inequality.
Any help will be appreciated.

Answer 1

The first inequality for $n=3$.

By AM-GM $$\sum_{cyc}\frac{a}{b}+\frac{9\sqrt[3]{abc}}{a+b+c}=\frac{1}{3}\sum_{cyc}\left(\frac{2a}{b}+\frac{b}{c}\right)+\frac{9\sqrt[3]{abc}}{a+b+c}\geq$$ $$\geq\sum_{cyc}\sqrt[3]{\frac{a^2}{b^2}\cdot\frac{b}{c}}+\frac{9\sqrt[3]{abc}}{a+b+c}=\sum_{cyc}\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq$$ $$\geq6\sqrt[6]{\left(\frac{a+b+c}{3\sqrt[3]{abc}}\right)^3\left(\frac{3\sqrt[3]{abc}}{a+b+c}\right)^3}=6.$$

Answer 2

A proof for $n=3$ of the second one.

By AM-GM $$\sum_{cyc}\frac{a^2}{b}+\frac{9}{a^2+b^2+c^2}\geq2\sqrt{\sum_{cyc}\frac{a^2}{b}\cdot\frac{9}{a^2+b^2+c^2}}.$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{a^2}{b}\geq a^2+b^2+c^2$$ or $$\sum_{cyc}\frac{a^2}{b}\sum_{cyc}ab\geq\sum_{cyc}a^2\sum_{cyc}a$$ or $$\sum_{cyc}(a^4c^2-a^3c^2b)\geq0,$$ which is true by AM-GM again: $$\sum_{cyc}a^4c^2=\frac{1}{6}\sum_{cyc}\left(4a^4c^2+b^4a^2+c^4b^2\right)\geq\sum_{cyc}\sqrt[6]{(a^4c^2)^4\cdot b^4a^2\cdot c^4b^2}=\sum_{cyc}a^3c^2b.$$