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What is the least possible value of $$ (x+1)(x+2)(x+3)(x+4)+2019 $$ where $x$ is a real number? 
Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation
Grouping the first and last terms and two middle terms gives $\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)+2019$, which can be simplified to $\left(x^{2}+5 x+5\right)^{2}-1+2019$. Noting that squares are nonnegative, and verifying that $x^{2}+5 x+5=0$ for some real $x$, the answer is 2018. 