1. ## Equality

For non-zero a,b,c

$\displaystyle a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$

show also that equality holds if and only if $\displaystyle a,b,c\in\{-1,1\}$.

How can I show this?

2. Originally Posted by dwsmith
For non-zero a,b,c
$\displaystyle a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$
show also that equality holds if and only if $\displaystyle a,b,c\in\{-1,1\}$.
That "only if" part is not true.
$\displaystyle \left( {\forall x,y} \right)\left[ {x^2 + y^2 \geqslant 2xy} \right]$.
If $\displaystyle t\ne 0\text{ let }x=t~\&~y=\frac{1}{t},~t=a,b,c$.

3. Originally Posted by dwsmith
For non-zero a,b,c

$\displaystyle a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$

show also that equality holds if and only if $\displaystyle a,b,c\in\{-1,1\}$.

How can I show this?
$\displaystyle a^2+\frac{1}{a^2}\geq2$

$\displaystyle b^2+\frac{1}{b^2}\geq2$

$\displaystyle c^2+\frac{1}{c^2}\geq2$

(Using AM-GM inequality three times)

$\displaystyle a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$
Each of the AM-GM inequalities becomes an equality if and only if $\displaystyle a^2=\frac{1}{a^2}$, $\displaystyle b^2=\frac{1}{b^2}$, $\displaystyle c^2=\frac{1}{c^2}$
ie $\displaystyle a=\pm1$, $\displaystyle b=\pm1$, $\displaystyle c=\pm1$
ie $\displaystyle a,b,c\in\{-1,1\}$