Something Fishy

Let $x = \tan a$ and $y = \tan b$ . Then, we have:

$\begin{aligned} Q & = \frac{(x+y)(1-xy)}{(1+x^2)(1+y^2)} \\ & = \frac{(\tan a + \tan b)(1 - \tan a \tan b)}{(1 + \tan^2 a)(1 + \tan^2 b)} \\ & = \frac{(\tan a + \tan b)(1 - \tan a \tan b)}{\sec^2 a \sec^2 b} \\ & = \cos^2 a \cos^2 b \left(\frac{\sin a}{\cos a} + \frac{\sin b}{\cos b} \right) \left(1 - \frac{\sin a \sin b}{\cos a \cos b} \right) \\ & = (\sin a \cos b + \sin b \cos a)(\cos a \cos b - \sin a \sin b) \\ & = \sin (a+b) \cos (a+b) \\ & = \frac{\sin (2(a+b))}{2} \end{aligned}$

$\Rightarrow \begin{cases} Q_{max} = \frac{1}{2} \\ Q_{min} = - \frac{1}{2} \end{cases} \quad \Rightarrow Q_{max}Q_{min} = - \frac{1}{4} = \boxed{-0.25}$

let $(x + y) = A$ and $(1 - xy) = B$

Then

$(1 + x^2)(1 + y^2) = 1 + x^2 + y^2 + x^2y^2$

$= 1 - 2xy + x^2y^2 + x^2 + 2xy + y^2$

$= (1 - xy)^2 + (x + y)^2$

$= A^2 + B^2$

So the expression can be express by

$\frac{AB}{A^2 + B^2}$

let the above expression as equal to $k$

then

$A^2 + \frac{1}{k} AB + B^2 = 0$

As x, y can only have real values, the discriminant should have value of 0 or above.

therefore

$1/k^2 - 4 \ge 0$

$(2k - 1)(2k + 1) \le 0$

$\therefore -1/2 \le k \le 1/2$

so the answer is $-0.5 \times 0.5 = -0.25$

Try this:

Let $P$ denote the expression above. Then $|P|=|\frac{(x+y)(1-xy)}{(x^2+1)(y^2+1)}|=\frac{|(x+y)(1-xy)|}{(x+y)^2+(1-xy)^2}$ (just expand the denominator).

Using AM-GM inequality we have $(x+y)^2+(1-xy)^2 \geq 2|(x+y)(1-xy)| \Rightarrow 0 \leq |P| \leq \dfrac 12$ .

Or $-\dfrac 12 \leq P \leq \dfrac 12$ . $P$ can reach $\dfrac 12$ when $(x;y)=(0;1)$ and reach $-\dfrac 12$ when $(x;y)=(0;-1)$ .

So the answer is $\dfrac 12 \times (-\dfrac 12) = -\dfrac 14 = \boxed{-0.25}$ .