Minimum of a complicated expression

Clearing out the denominators, we see that the expression equals $\frac{a^4+b^4+c^4+a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}$ Applying AM-GM inequality on the set $(a^4, b^2c^2)$ , we find out $a^4+b^2c^2 \geq 2a^2bc$ Similarly, $b^4+c^2a^2 \geq 2b^2ca$ $c^4+a^2b^2 \geq 2c^2ab$ Summing them up, $a^4+b^4+c^4 + a^2b^2+b^2c^2+c^2a^2\geq 2(a^2bc+b^2ca+c^2ab)$ Again, from AM-GM inequality, $a^2bc+b^2ca+c^2ab \geq 3 \sqrt[3]{a^3b^3c^3}= 3abc$ So $a^4+b^4+c^4+a^2+b^2+c^2 \geq 6abc$ $\frac{a^4+b^4+c^4+a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2} \geq 6$ Equality holds for $(a, b, c)= (1, 1, 1)$ . We thus conclude our answer is $\boxed{6}$ .

I solved this at first glance by simple observation. The terms in the expression are very neatly cyclic. So, if we want to take advantage by increasing $a$ , then decreasing $b$ and bla bla bla, it's actually disadvantageous since the variables $a,b,c$ are cyclically distributed in the whole expression. For example, if we want to minimize $\dfrac{b^3}{ca}$ and so we minimize $b$ , then maximize $ca$ , we are in another way maximizing $\dfrac{ca}{b}$ . Hence, maximizing, minimizing, increasing, decreasing doesn't actually help.

Hence, seeing the question first, the only solution I could think of is that $a=b=c$ , implying the minimum value to be $\fbox{6}$ , and it worked. :p

Sorry for being less mathematical, more logical... :D