Equality

**dwsmith** · May 21st 2011, 08:54 AM

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?

**Plato** · May 21st 2011, 09:13 AM

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 $t\ne 0\text{ let }x=t~\&~y=\frac{1}{t},~t=a,b,c$ .

**alexmahone** · May 21st 2011, 09:20 AM

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)

Adding,

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

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 $a,b,c\in\{-1,1\}$

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