Maximum of Roots!

Let $y$ denote the value of this expression. Then the turning point of $y$ occurs when $\dfrac{dy}{dx} = 0$ ,

$\dfrac{-1}{2\sqrt{2008 - x}} + \dfrac1{2\sqrt{x-2000}} = 0 \quad \Leftrightarrow \quad 2008 - x= x-2000 \quad \Leftrightarrow \quad x_m = x = 2004 = 2^2\times3\times167 .$

A simple use of the second derivative test confirms that this turning point is a maximum point.

The answer is T H R E E .

First, Let $y = \sqrt{2008-x} + \sqrt{x-2000}$ .

$\begin{aligned} y^2 & = \left( \sqrt{2008-x} + \sqrt{x-2000} \right)^2 \\ & = 2008 - x + x - 2000 + 2\sqrt{(2008-x)(x-2000)} \\ & = 8 + 2\sqrt{(2008-x)(x-2000)} \end{aligned}$

To maximize $y$ , then $(2008-x)(x-2000)$ must be a perfect square. Then, we can clearly see that if $(2008-x)(x-2000)$ is a perfect square, then $2008 - x = x - 2000$ . Hence, $x = 2004$ .

So, $y$ has the maximum value at $x_m = x = 2004$ .

We note that $2004 = 2^2 \times 3 \times 167.$

And, hence the answer is $3$

Using Cauchy-Schwarz inequality :

$\begin{aligned} \left(\sqrt{2008-x} + \sqrt{x-2000}\right)^2 & \le 2(2008-x+x-2000) = 16 \\ \implies \sqrt{2008-x} + \sqrt{x-2000} & \le 4 \end{aligned}$

Equality occurs when $\sqrt{2008-x} = \sqrt{x-2000} \implies x = 2004 = 2^2\times 3 \times 167$ , that is $\boxed{3}$ distinct prime factors.