(x+3)^4=x^4+12x^3+54x^2+108x+81; (x-1)^4=x^4-4x^3+6x^2-4x+1.
Add these: 2x^4+8x^3+60x^2+104x+82 and this expression ≥0.
So x^4+4x^3+30x^2+52x≥0 because the 82s cancel out and the expression is divisible by 2.
Therefore x(x^3+4x^2+30x+52)≥0.
x+2 is a factor of the cubic: x(x+2)(x^2+2x+26)≥0. The quadratic has only complex solutions.
-2 | 1 4 30 52
1 -2 -4 -52
1 2 26 | 0
Therefore x≤-2 satisfies the inequality: try x=-3: (-3)(-1)(29)≥0 is true.
When x=0 or -2 the inequality holds because the expression=0.
And x≥0 satisfies the inequality: try x=1: 1(3)(29)≥0 is true.
SOLUTION x≤-2 or x≥0. In the title of the question > appears; but in the text it is >=. Adjust the solution according to which inequality is intended.