To prove or disprove $abc=1$ $\implies$ $ \sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4} > \dfrac92$ Let $a,b,c$ be positive real numbers such that $abc=1$ , then is it true that
$ \sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4} > \dfrac92$ ? 

With the hope that the problem remains unchanged; 

First see that:
$$
\sqrt{a^4+a^2b^2+b^4}\geq \sqrt 3ab
$$
Hence:
$$
\sqrt {a^4+a^2b^2+b^4}+\sqrt {b^4+b^2c^2+c^4}+\sqrt {c^4+c^2a^2+a^4}\\\geq \sqrt 3(ab+bc+ac)\geq 3\sqrt 3> \dfrac92
$$
