prove that $\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$ prove that
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$$
where $a,b,c>0$ and $n\ge0,n\le 3$,$a+b+c=ab+bc+ca$
My try: by given condition $a+b+c=ab+bc+ca$
we have $a+b+c\le a^2+b^2+c^2$
also using titu's lemma
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge a+b+c +\frac{3n}{a^2+b^2+c^2}$$
or
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge ab+bc+ca +\frac{3n}{a^2+b^2+c^2}$$
i dont know what to do next. Any ideas preferably using am-gm.
source Samin Riasat Basics in Olympiad ineq.

We can use also, the following much more weaker estimation:
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{(a^2+b^2+c^2)(a+b+c)}{ab+ac+bc},$$ which is just $$\sum_{cyc}(a^2c-b^2a)^2\geq0.$$
Thus, it's enough to prove that:
$$\frac{(a^2+b^2+c^2)(a+b+c)}{ab+ac+bc}+\frac{9(ab+ac+bc)^3}{(a^2+b^2+c^2)(a+b+c)^3}\geq\frac{6(ab+ac+bc)}{a+b+c}.$$
Now, let $a^2+b^2+c^2=k(ab+ac+bc)$ again.
Thus, we need to prove that:
$$k+\frac{9}{k(k+2)^2}\geq\frac{6}{k+2},$$ which is true by AM-GM:
$$k+\frac{9}{k(k+2)^2}\geq2\sqrt{k\cdot\frac{9}{k(k+2)^2}}=\frac{6}{k+2}.$$
