You need to use Vieta's formulas to evaluate the summation and the product of the solutions to quadratic equation `x^2-2mx+m^2-1=0` , such that:
`ab = (m^2-1)/1 => ab = m^2-1`
`a + b = -(-2m)/1 => a + b = 2m`
Replacing `m^2-1` for `ab` and `2m` for `a + b` in the inequality, yields:
`m^2 - 1 - 2m + 2 >= 0`
`m^2 - 2m + 1 >= 0`
You need to notice that the quadratic `m^2 - 2m + 1` expression represents expansion of the square `(m - 1)^2,` such that:
`(m - 1)^2 >= 0`
Since a square is positive for all m and for `m = 1` , `m - 1 = 0` , yields that the inequality holds.
Hence, testing if the inequality `m^2 - 2m + 1 >= 0` holds, under the given conditions, yields that the statement `m^2 - 2m + 1 >= 0` is valid.