An atypical inequality

Question

If $a, b, c \in (0, \infty)$ with $abc=1$ prove: $$a^2+b^2+c^2+3(ab+bc+ca)\leq\frac{1}{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})(a+b)(b+c)(c+a).$$

On atypical inequality (a smaller amount than a product) that resisted my attempts to prove it.

Answer 1

We need to prove that $$\prod_{cyc}(a^2+b^2)\sum_{cyc}a\geq2abc\sum_{cyc}(a^4+3a^2b^2)$$ for positives $a$, $b$ and $c$ or $$\sum_{cyc}\left(a^4b^2+a^4c^2+\frac{2}{3}a^2b^2c^2\right)\sum_{cyc}a\geq2\sum_{cyc}(a^5bc+3a^3b^3c)$$ or $$\sum_{cyc}(a^5b^2+a^5c^2+a^4b^3+a^4c^3+a^4b^2c+a^4c^2b+2a^3b^2c^2-2a^5bc-6a^3b^3c)\geq0,$$ which is true because $$\sum_{cyc}(a^5b^2+a^5c^2-2a^5bc)=\sum_{cyc}a^5(b-c)^2\geq0$$ and by AM-GM $$\sum_{cyc}(a^4b^3+a^4c^3+a^4b^2c+a^4c^2b+2a^3b^2c^2-6a^3b^3c)=$$ $$=\sum_{cyc}(a^4b^3+a^3b^4+a^4b^2c+b^4a^2c+a^3b^2c^2+b^3a^2c^2-6a^3b^3c)\geq$$ $$\geq\sum_{cyc}\left(6\sqrt[6]{a^4b^3\cdot a^3b^4\cdot a^4b^2c\cdot b^4a^2c\cdot a^3b^2c^2\cdot b^3a^2c^2}-6a^3b^3c\right)=0.$$ Done!