Solve `3x^2-1 >= (3+x^2)/2` .
First we rewrite as an equivalent inequality (one that has the same solutions as the original)
`3x^2-1 >= (3+x^2)/2`
`6x^2-2>=3+x^2` Multiply both sides by 2
`5x^2-5>=0` Pull terms to left side by subtraction
`5(x^2-1)>=0 => 5(x+1)(x-1)>=0` Factor.
Now the left hand side is zero when x=1 or -1. To find the intervals where it is greater than zero we plug in test values. The intervals we are interested in are x<-1, -1<x<1, and x>1; so we try x=-2,x=0, and x=2.
If x=-2 the left hand side is positive; so if x<-1 then the left-hand side is greater than zero.
If x=0 the left-hand side is negative so the left-hand side is negative for -1<x<1.
If x=2 the LHS is positive so LHS is greater than zero if x>1.
We can also look at the graph to check our work -- notice that the "skinnier" parabola is above the wider parabola as long as x<-1 or x>1.
Thus the solution to `3x^2-1 >= (3+x^2)/2` is `x<= -1` or `x>=1` .