Help me prove this inequality for any positive numbers a,b,c

Question

Prove the inequality below for any positive numbers $a, b, c$ $$\frac{3+a^4+b^3+c^2}{1+2a^3+3b^2+6c}+\frac{3+b^4+c^3+a^2}{1+2b^3+3c^2+6a}+\frac{3+c^4+a^3+b^2}{1+2c^3+3a^2+6b}\geqslant\frac{3}{2}$$ I tried reducing the left side to a common denominator but it didn't help at all. Also I was told that this question is a "one shot" so maybe one of you guys will be able to notice that one thing that would help me prove this inequality.

Answer 1

By AM-GM $$3b^2\leq2b^3+1$$ and $$6c\leq2c^3+4.$$ Thus, $$\sum_{cyc}\frac{3+a^4+b^3+c^2}{1+2a^3+3b^2+6c}\geq\sum_{cyc}\frac{3+a^4+b^3+c^2}{1+2a^3+2b^3+1+2c^3+4}=\frac{\sum\limits_{cyc}(3+a^4+a^3+a^2)}{6+2(a^3+b^3+c^3)}.$$ Id est, it's enough to prove that $$\sum\limits_{cyc}(3+a^4+a^3+a^2)\geq3(3+a^3+b^3+c^3)$$ or $$\sum_{cyc}(a^4-2a^3+a^2)\geq0$$ or $$\sum_{cyc}a^2(a-1)^2\geq0$$ and we are done!