$\displaystyle |3x+1|>\sqrt{x+1}$

My working :

There will be 2 cases here .

Case 1 : $\displaystyle 3x+1>\sqrt{x+1}$

$\displaystyle 9x^2+6x+1>x+1$

$\displaystyle x(9x+5)>0$

Case 2 : $\displaystyle 3x+1<-\sqrt{x+1}$

$\displaystyle (3x+1)^2>x+1$

$\displaystyle x(9x+5)>0$ , Isn't this the same as case 1 ..

I think i made a mistake at case 2 . Can someone check for me .. Thanks .