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.

$a^2+b^2+c^2+a^{-2}+b^{-2}+c^{-2}=(a-a^{-1})^2+(b-b^{-1})^2+(c-c^{-1})^2+6$, whence the minimum occurs when $a=a^{-1},b=b^{-1},c=c^{-1}$ and is $6$.
