Given an absolute value inequality, find all values of a such that it will satisfy the equation for all real numbers

Question

Suppose I have an absolute value inequality:

$x^2 -$ $|x - a|$ $-$ $|x - 1|$ $+$ $3 ≥ 0$

What strategy should I use to find “$a$” in this case?

Answer 1

This is not a complete solution. It is meant as a possible guide.

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Let $f(x) = x^2 - |x-1 | + 3 = \begin{cases} x^2 - x + 4 & x\geq 1 \\ x^2 + x + 2 & x < 1 \\ \end{cases}$

We want to find $ f(x) \geq |x-a |$.
A look at the graph of $f(x)$ suggests that it's the negative values of $x$ that are more likely to violate the conditions.

If $ a > 1$, then at $ x = -1, f(-1) = 1 -1 + 2 = 2$ and $|-1 -a | = a+1 > 2$, so there are no solutions.
However, this is quite contrived (IE how do we know to use $x = -1$ to eliminate all cases?) so is there another approach here?

What happens when $ a \leq 1$?

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Note: If you know calculus, then consider what happens at the solution to $ f(x) = | x-a|$ given that $f(x) \geq |x-a|$. What can we say about $f'(x^*)$?