The minimum value of $a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}?$

Question

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.

Answer 1

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

Answer 2

Might as well take advantage of the fact that it's a multiple choice question.

First, is it possible that the quantity is ever zero? Next, can you find $a, b, c$ such that $$a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} = 6?$$

Answer 3

By the symmetry of the expression, the minimum is attained with $a=b=c$ so we look for the minimum of $$f(x)=3(x^2+x^{-2})$$ By the derivative of the function $f$ we find that the minimum is at $x=1$, hence your answer must be $6$.

Answer 4

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

$\sum (a_i-\dfrac{1}{a_i})^2 \ge 0$

$\sum (a_i)^2+\dfrac{1}{(a_i)^2} \ge 2 \cdot i$