# Equality

• May 21st 2011, 09:54 AM
dwsmith
Equality
For non-zero a,b,c

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

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

How can I show this?
• May 21st 2011, 10:13 AM
Plato
Quote:

Originally Posted by dwsmith
For non-zero a,b,c
$a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$
show also that equality holds if and only if $a,b,c\in\{-1,1\}$.

That "only if" part is not true.
$\left( {\forall x,y} \right)\left[ {x^2 + y^2 \geqslant 2xy} \right]$.
If $t\ne 0\text{ let }x=t~\&~y=\frac{1}{t},~t=a,b,c$.
• May 21st 2011, 10:20 AM
alexmahone
Quote:

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

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

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

How can I show this?

$a^2+\frac{1}{a^2}\geq2$

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

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

(Using AM-GM inequality three times)

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