Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$
Originally I thought to rephrase it in geometric terms, seeing one degree terms I think about equation of a line.
I also have that $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leq p x+q \leq \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}$
I think that a geometric solution might be possible, please help me to proceed.
Thanks.
Best Answer
Let $f(x) = \sqrt{1-x^{2}}-p x-q$, then the condition becomes $$|f(x)| \leq \frac{\sqrt{2}-1}{2}$$ Thus, the extrema of $f$ should lie in $\left[ -\frac{\sqrt{2}-1}{2}, \frac{\sqrt{2}-1}{2}\right]$. Now, let's see how its extrema look like: $$f'(x) = \frac{-x}{\sqrt{1-x^2}} - p = 0 $$ After bringing to a common denominator, numerator should be equal to zero: $$ -x - p\sqrt{1-x^2} = 0 \\[2mm] x = r\sqrt{1-x^2}, \ \ (r=-p) \\[2mm] x^2 = r^2(1-x^2) \\[2mm] (1+r^2)x^2 = r^2 \\[2mm] x_0 = \pm \frac{r}{1+r^2} = \mp\frac{p}{\sqrt{1+p^2}}$$ However, since $x \ge 0$, we get $$x_0 = -\frac{p}{\sqrt{1+p^2}} $$ Therefore, $$\begin{align}f(x_0) &= \sqrt{1-\frac{p^2}{1+p^2}} + \frac{p^2}{\sqrt{1+p^2}}-q \\[1mm]&= \frac{1 + p^2}{\sqrt{1+p^2}} -q \\[1mm]&= \sqrt{1+p^2}-q \end{align}$$ Hence, the condition seems to be satisfied for all $p$ and $q$ such that $$\left| \sqrt{1+p^2} -q \right| \le \frac{\sqrt2 - 1}{2} \tag{1}$$ Also, we need to see the edge cases when $x = 0$ and $x = 1$, which give $$|f(0)| = \left|1 - q\right| \le \frac{\sqrt{2}-1}{2} \tag{2}$$ and $$|f(1)| = \left|-p - q\right| \le \frac{\sqrt{2}-1}{2} \tag{3}$$ Finally, combining $(1)$, $(2)$ and $(3)$ we get the final answer.
I didn't solve the last three inequalities putting all together, but after graphing I noticed (but still needs to be justified rigorously by solving the inequalities) that only one pair satisfies the statement, namely, $$(p,q) = \left(-1, \frac{\sqrt 2 + 1}{2}\right)$$
Below is the graph with those $p$ and $q$: