# Thread: can't figure this inequality

1. ## can't figure this inequality

$(M^2+2MN-N^2 )^2<(M^2-2MN-N^2 )^2$

I went about it like this

$M^2+2MN-N^2
$M^2+2MN-N^2>-M^2+2MN+N^2\ \ \ \ \ \ (2)$

$(1)$

$M^2+2MN-N^2

$2MN<-2MN$

$\implies\ \ \ \ \ M<00>N$

$(2)$

$M^2+2MN-N^2>-M^2+2MN+N^2$

$M^2-N^2>-M^2+N^2$

$2M^2-2N^2>0$

$(M+N)(M-N)>0$

so $(M+N)$ and $(M-N)$ have the same sign

$(M+N)\ \ and\ \ (M-N)\ \ >0\ \ \ \ (3)$

$(M+N)\ \ and\ \ (M-N)\ \ <0\ \ \ \ (4)$

$(3)$

$(M+N)\ \ and\ \ (M-N)\ \ >0$

$\implies\ \ \ \ \ \ M>|N|$

$(4)$

$(M+N)\ \ and\ \ (M-N)\ \ <0$

$\implies\ \ \ \ \ \ M<|N|$

so altogether I have

$M>0>N$
$M<0
$M>|N|$
$M<|N|$

which seems to imply that all M and N fit apart from M=N

let M=2 N=1

$(M^2+2MN-N^2 )^2=7\ \ \ \ \(M^2-2MN-N^2 )^2=1$

but this has $(M^2+2MN-N^2 )^2>(M^2-2MN-N^2 )^2$

What am I doing wrong ?

2mn < -2mn
1 < -1

stop!

2mn<-2mn

m=2
n=-1
2mn=-4
-2mn=4
2mn<-2mn

4. ## Re: can't figure this inequality

Multiply out both sides and collect terms and you arrive at

$MN(M^{2}-N^{2}) < 0$

which clearly is not the case for all $M$ and $N.$

If $M$ and $N$ are both positive you would need to have $M

5. ## Re: can't figure this inequality

Thanks Bob, seems easier than the way I was going about it.

$MN(M^2-N^2)<0$

$M\ne0$

$M>0\ \ \ \implies \ \ \ 0

$M<0\ \ \ \implies \ \ \ NN>0$

Does this cover all of them ?