# 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.

• Note that the function is a sum $f(a) + f(b) + f(c)$, so you can optimize in $a, b, c$ separately since the maximum value is obtained when each of the individual terms are maximized. – Qiaochu Yuan Mar 29 '13 at 8:39
• @QiaochuYuan 's comment continued regarding separate optimisation: $\frac1u +u\geqslant 2\sqrt\frac uu=2\,$ for every $u>0$, and "$=$" only holds for $u=1$. All that is due to the AGM inquality. – Hanno Feb 12 '19 at 13:23

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

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 $$a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} = 6?$$

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 $6$.

• But this doesn't prove that the minimum exists. (Although it is irrelevant in a multiple choice question.) – Bartek Mar 29 '13 at 3:53
• @Bartek I proved that if there's a minimum then it must be $6$. Now the expression is positive so the set $\{a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}, a,b,c\not=0\}$ is not empty and bounded below by $0$ then it has an infimum which's the minimum already founded – user63181 Mar 29 '13 at 4:05
• Why is the infimum a minimum? Doesn't it need proving that the infimum is attained? – Bartek Mar 29 '13 at 11:19
• @Bartek By the same proof given in my answer, do you not agree that the infinimum of the expression in initial problem is equal to $\inf_{x\not=0}f(x)=6=f(1)$, so the infimum is attained? – user63181 Mar 29 '13 at 11:37
• No. You proved that if a minimum exists, then it is equal to $6$ and is attained in $1$. You also show in a comment that an infimum exists. But the existence of an infimum doesn't in general imply the existence of a minimum. – Bartek Mar 29 '13 at 11:41

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

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

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

• Thanks a lot for the nice method. +1 from me. – user52976 Mar 29 '13 at 11:50