<s>[INST] For positive $a$, $b$, $c$ with $a^4+b^4+c^4=3$, then $\sum_{cyc}\frac{a^2}{b^3+1}\geq \frac32$ 
If $a$, $b$, $c$ $\in (0, \infty)$,  $a^4+b^4+c^4=3$, then: $$\sum_{cyc}\frac{a^2}{b^3+1}\geq \frac32$$

original problem image
I have been into inequalities lately and I am stuck with this. I used a famous inequality at first 
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3 \left(\frac{a^4+b^4+c^4}{3}\right)^{1/4}$$ 
From this, I just had to prove that $x^3+1-2x \le 0$ for all $x \in (0,1)$ but infact this is not true. 
Can anyone help me out? 
 [/INST] Note that
$$\frac{1}{b^3+1} - \Big(1 - \frac{b^2+b}{4}\Big) = \frac{b(b^2+3b+1)(b-1)^2}{4(b+1)(b^2-b+1)}\ge 0.$$
It suffices to prove that
$$\sum_{\mathrm{cyc}} a^2\Big(1 - \frac{b^2+b}{4}\Big) \ge \frac{3}{2}$$
or
$$a^2+b^2+c^2 - \frac{1}{4}(a^2b^2+b^2c^2+c^2a^2) - \frac{1}{4}(a^2b+b^2c+c^2a)\ge \frac{3}{2}.$$
Note that
$$a^2b+b^2c+c^2a \le \sqrt{(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)}.$$
It suffices to prove that
$$a^2+b^2+c^2 - \frac{1}{4}(a^2b^2+b^2c^2+c^2a^2) - \frac{1}{4} \sqrt{(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)} \ge \frac{3}{2}.$$
Let $p = a^2+b^2+c^2, \ q = a^2b^2 + b^2c^2 + c^2a^2$.
It follows from $a^4+b^4+c^4 = 3$ that $p^2 - 2q = 3$ or $q = \frac{p^2 - 3}{2}.$
It is easy to obtain $\sqrt{3}\le p \le 3.$
We need to prove that 
$$p - \frac{1}{4}q - \frac{1}{4}\sqrt{pq} \ge \frac{3}{2}$$
or
$$8p-p^2-9 \ge \sqrt{2p(p^2-3)}.$$
Clearly, $8p-p^2-9 > 0$. It suffices to prove that $$(8p-p^2-9)^2 - 2p(p^2-3)\ge 0$$
or $$(p-3)(p^3-15p^2+37p-27)\ge 0.$$ It is obvious. We are done.
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