2019 AMC 10 A

Grouping the first and last terms and two middle terms gives , which can be simplified to . Noting that squares are nonnegative, and verifying that for some real , the answer is .

Let . Then the expression becomes .

We can now use the difference of two squares to get , and expand this to get . Refactor this by completing the square to get , which has a minimum value of . The answer is thus .

Similar to Solution , grouping the first and last terms and the middle terms, we get .

Letting , we get the expression . Now, we can find the critical points of to minimize the function:

To minimize the result, we use . Hence, the minimum is , so .

Note: We could also have used the result that minimum/maximum point of a parabola occurs at .

The expression is negative when an odd number of the factors are negative. This happens when or . Plugging in or yields , which is very close to . Thus the answer is .

Using the answer choices, we see that choices , , and are impossible,since can actually be negative (as seen when e.g.). Plug in to see that it becomes , so round this to .