Problem 1439

We will show that the maximum value is $\frac{1}{2}$ and the minimum is $-\frac{1}{2}$. For maximum, after some rearranging, we want to prove
$$2(ad + bc - cd) \leq c^2 + d^2,$$
or
$$2(ad + bc) \leq (c + d)^2.$$
After adding double the expression $ac + bd = a^2 + b^2$ to both sides of this inequality, we will get equivalent inequality
$$2(a + b)(c + d) = 2(ad + bc + ac + bd) \leq (c + d)^2 + 2a^2 + 2b^2,$$
which can be further rearranged to
$$0 \leq (c+d)^2 - 2(a+b)(c+d) + (a+b)^2 + (a-b)^2 \leq (c+d-a-b)^2 + (a-b)^2,$$
which clearly holds. Moreover, this maximal value is reached by $a = b = c = d > 0$.

For the minimum value, notice that if $a \geq c$ or $b \geq d$ holds, $ad + bc \geq ad \geq cd$ and the expression is non-negative. So for it to be negative, both $a < c$, $b < d$ must hold and $a^2 + b^2 < ac + bd$ if $a$ and $b$ wouldn't both be $0$. As they are equal by given condition, indeed $a = b = 0$ and the minimized expression then becomes $-\frac{cd}{c^2+d^2}$, which has minimum $-\frac{1}{2}$ as $2cd \leq c^2 + d^2$. Moreover, this minimal value is reached by $a = b = 0$ and $c = d > 0$.

Consider the Cartesian coordinate system and in it, points $C = (c, 0)$, $D = (0, d)$ and $X = (a, b)$. The line passing through $C$, $D$ has equation $dx + cy - cd = 0$. From analytic geometry, the formula for distance of point $(m, n)$ to the line $ix + jy + k = 0$ is known. It is $\frac{im+jn+k}{\sqrt{i^2+j^2}}$, where the distance is oriented according to the vertical position of point $(m, n)$ with respect to the given line. With this formula, the distance of point $X$ to the line through $C$, $D$ is
$$\frac{ad + bc - cd}{\sqrt{c^2 + d^2}}.$$
By rearranging the given condition on $a$, $b$, $c$, $d$, we get
$$\left(a - \frac{c}{2}\right)^2 + \left(b - \frac{d}{2}\right)^2 = \frac{c^2 + d^2}{4},$$
which means that the point $X = (a, b)$ lies on the circle with centre $(\frac{c}{2}, \frac{d}{2})$ and radius $\frac{\sqrt{c^2+d^2}}{2}$, which is exactly the circle with diameter $CD$. Such point on circle with diameter $CD$ can be at most radius distant from $AB$, so
$$\frac{ad + bc - cd}{\sqrt{c^2 + d^2}} \leq \frac{\sqrt{c^2 + d^2}}{2}$$
and by rearranging we get the inequality we proved in first solution. Note that the distance from $X$ to $CD$ is nonnegative unless $X = (0,0)$, as $X$ would be above the line $CD$, because it lies in the first quadrant by the nonnegativity and the halfcircle with diameter $CD$ in first quadrant lies entirely above the line $CD$. From this, the minimum value must happen for $X = (0,0)$ which gives $-\frac{1}{2}$, as desired.