If x is any real number, then prove that $\displaystyle x^2+2x+4\geq0$.
alternatively, you could look at the discriminant: b^{2} - 4ac of the quadratic x^{2}+2x + 4, which is:
4 - 4(1)(4) = -12 < 0.
this tells you that the parabola:
y = x^{2}+2x + 4 never crosses the x-axis, and since it "opens up" (the x^{2} term is positive), it must lie entirely above the x-axis
(not only is the inequality true, but we can also replace "≥" with ">". in fact, the value of x^{2}+2x + 4 never gets smaller than 3, which happens at x = -1).