SAT1000 - P527

$\dfrac 1{2|a|} + \dfrac {|a|}b = \dfrac 1{2|2-b|} + \dfrac {|2-b|}b = \begin{cases} \dfrac 1{2(2-b)} + \dfrac {2-b}b & \text{for }0 < b \le 2 \\ \dfrac 1{2(b-2)} + \dfrac {b-2}b & \text{for }b > 2 \end{cases}$

Case 1: $0 < b \le 2$

$\begin{aligned} f(b) & = \frac 1{2(2-b)} + \frac {2-b}b \\ f'(b) & = \frac 1{2(2-b)^2} - \frac 2{b^2} = - \frac {3b^2-16b+16}{2(2-b)^2b^2} = - \frac {(3b-4)(b-4)}{2(2-b)^2b^2} \\ \min (f(b)) & = f \left(\frac 43 \right) = \frac 54 \end{aligned}$

Case 2: $b > 2$

$\begin{aligned} f(b) & = \frac 1{2(b-2)} + \frac {b-2}b \\ f'(b) & = - \frac 1{2(b-2)^2} + \frac 2{b^2}= \frac {(3b-4)(b-4)}{2(2-b)^2b^2} \\ \min (f(b)) & = f (4) = \frac 34 \end{aligned}$

Therefore the minimum value $M=\frac 34 = 0.75$ , $\implies \lfloor 1000M \rfloor = \boxed{750}$ .

This problem can be solved using calculus easily. Firstly, we can turn the expression $\frac{1}{2|a|} + \frac{|a|}{b}$ into a single variable function using the condition $a + b = 2$ . Let $f(b) = \frac{1}{2|2 - b|} + \frac{|2 - b|}{b}$

We now take the derivative of $f(b)$ . This results in the piecewise function $f'(b) = \frac{2}{\left(4-2b\right)^{2}}-\frac{2}{b^{2}}$ if b < 2 and $-\frac{2}{\left(2b-4\right)^{2}}+\frac{2}{b^{2}}$ if b > 2 with a discontinuity at b = 2. Since we are trying to find the minimum of $f(b)$ , we must find the critical points of $f(b)$ . It just so happens that in both halves this occurs when $b^{2}=\left(4-2b\right)^{2}$ . Rearranging terms yields the quadratic equation $3b^{2} - 16b + 16 = 0$ which has the roots $b = 1.333, 4$ .

Since we are given that there exists a global minimum, one of these critical points must be the global minimum (under different circumstances I would have to show that one of these is a global minimum, most likely via a second-derivative test). Plugging both roots into $f(b)$ yields the local minimums (1.333, 1.25) and (4, 0.75). It is clear that M = 0.75 meaning that $\lfloor 1000M \rfloor = 750$

Let the given sum be $S$ . Then

$\displaystyle lim_{a \to -\infty } S=1,\displaystyle lim_{a \to 0^{-}} S=\infty$

In the range $(-\infty, 0)$ , it has one minimum at $a=-2,b=4$ , the minimum value being $0.75$ .

Also, $\displaystyle lim_{a \to 0^{+}} S=\displaystyle lim_{a \to 2^{-}} S=\infty$ .

Since $b>0, a<2$

In the range $[0,\infty), s$ has one minimum at $a=\frac{2}{3}, b=\frac{4}{3}$ , the minimum value being $1.25>0.75$ .

Hence the given sum can never be less than $\boxed {0.75}$ .