<s>[INST] Strange Cube Root Offense in an Inequality 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.
 [/INST] 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:
$$\sum_{cyc}a^4c^2=\frac{1}{6}\sum_{cyc}\left(4a^4c^2+b^4a^2+c^4b^2\right)\geq\sum_{cyc}\sqrt[6]{(a^4c^2)^4\cdot b^4a^2\cdot c^4b^2}=\sum_{cyc}a^3c^2b.$$
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