<s>[INST] 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.
 [/INST] We can use the following estimation.
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}.$$
The proof you can see here: https://artofproblemsolving.com/community/c6h296853
For hardest case $n=3$ we need to prove that:
$$\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}+\frac{9}{a^2+b^2+c^2}\geq6$$ or
$$\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}+\frac{9(ab+ac+bc)^3}{(a^2+b^2+c^2)(a+b+c)^3}\geq\frac{6(ab+ac+bc)}{a+b+c}.$$
Now, let $a^2+b^2+c^2=k(ab+ac+bc).$
Thus, $k\geq1$ and we need to prove that:
$$\frac{37k-19}{6}+\frac{9}{k(k+2)}\geq6$$ or
$$37k^4+74k^3-55k^2-110k+54\geq0$$ or
$$37k^4-37k^3+111k^3-111k^2+56k^2-56k-54k+54\geq0$$ or
$$(k-1)(37k^3+111k^2+56k-54)\geq0,$$ which is obvious.
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