Question

What is the least possible value of (x+1)(x+2)(x+3)(x+4)+2019 where x is a real number?

Answer 1

By completing the square, we have

`(x+1)(x+4) = x^2 + 5x + 4 = \left(x+\frac52\right)^2 - \frac94`

`(x+2)(x+3) = x^2 + 5x + 6 = \left(x + \frac52\right)^2 - \frac14`

Then

`(x+1)(x+2)(x+3)(x+4) = \left(x+\frac52\right)^4 - \frac52 \underbrace{\left(x + \frac52\right)^2}_(=y) + \frac9{16} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = y^2 - \frac52 y + \frac9{16} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \left(y - \frac54\right)^2 - 1 \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \left(\left(x + \frac52\right)^2 - \frac54\right)^2 - 1`

This product has a minimum of -1 when

`\left(x + \frac52\right)^2 - \frac54 = 0`

which has two real solutions; then the minimum of the overall expression is -1 + 2019 = 2018.