solve for m x^2+mx+1/x^2+x+1<3 for x=R?

Answer 1

x^2+mx+1/x^2+x+1<3

case 1 : for x>0
=> (x+1/x)^2 +(m+1)x <4
=> (m+1)x < 4- (x+1/x)^2

Now, we find the maximum value of the RHS:
the minimum value of (x+1/x)^2 is 2 when x=1 and for all other values of x it is >2
This makes max(RHS)=0 or m+1 <0
or m<-1

case 2: x<0
the same argument as above(but x=-1) leads us to m+1 >0
or m>-1

case 3: x=0
The given expression is undefined

I guess its a home work problem, which indeed makes you lazy!

Answer 2

m<(2-x-(1/x^2)-x^2)/x

Assuming x>0

m < (2/x)-1-(1/x^3)-x

Answer 3

solve it your self....lazy!!