Prove this inequality $\sum_{cyc}\sqrt{\frac{a}{3b^2+1}}\ge \frac{3}{2}$

Question

Let $a>0$, $b>0$ and $c>0$ such that $a+b+c=3$. Prove that: $$\sqrt{\frac{a}{3b^2+1}}+\sqrt{\frac{b}{3c^2+1}}+\sqrt{\frac{c}{3a^2+1}}\ge \frac{3}{2}$$

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I tried the AM-GM, Holder, C-S inequalities but was unsuccessful.

Answer 1

Holder helps!

$$\left(\sum_{cyc}\sqrt{\frac{a}{3b^2+1}}\right)^2\sum_{cyc}a^2(3b^2+1)(a+b+3c)^3\geq$$ $$\geq\left(\sum_{cyc}a(a+b+3c)\right)^3=\left(\sum_{cyc}(a^2+4ab)\right)^3.$$ Thus, it remains to prove that $$4\left(\sum_{cyc}(a^2+4ab)\right)^3\geq9\sum_{cyc}a^2(3b^2+1)(a+b+3c)^3,$$ which after homogenization and full expanding gives $$\sum_{cyc}(a^6+36a^5b+18a^5c+24a^4b^2+12a^4c^2+16a^3b^3+$$ $$+342a^4bc+348a^3b^2c+354a^3c^2b-1151a^2b^2c^2)\geq0,$$ which is obvious by AM-GM.

Done!