The minimum value of $a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}?$ I came across the following problem :  

Let $a,b,c$ are non-zero real numbers .Then the minimum value of $a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}?$ This is a multiple choice question and the options are $0,6,3^2,6^2.$  

I do not know how to progress with the problem. Can someone point me in the right direction? Thanks in advance for your time.

Did you try $AM-GM$?
$\dfrac{(a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})}{6} \ge 1$
$a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \ge 6$
OR
$\sum (a_i-\dfrac{1}{a_i})^2  \ge 0$
$\sum (a_i)^2+\dfrac{1}{(a_i)^2} \ge 2 \cdot i$
