If $a$, $b$, $c$ $\in (0, \infty)$, $a^4+b^4+c^4=3$, then: $$\sum_{cyc}\frac{a^2}{b^3+1}\geq \frac32$$
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?
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.
