Find the minimum value of

Computing the derivative of the function yields a quartic polynomial, with no easy way of finding solutions (ugly coefficients). Wishful thinking guides us to conjecture an elementary solution instead. The best thing that could happen to us is if the function could be expressed as a product or sum of perfect squares. Noticing the $40x = 2 \times 40x$ , let's introduce a $(x+20)^2$ term to write the function as $(x+20)^2 + \frac{90000}{x^2} + \frac{9000}{x} + 225$ .

Now notice that $300^2 = 90000$ so let's consider introducing a term of the form $\frac{300}{x}$ . $\frac{90000}{x^2} + \frac{9000}{x} + 225 = \left(\frac{300}{x}\right)^2 + 30 \frac{300}{x} + 225$ . Finally, noticing that $225 = 15^2$ allows us to write all of this as $\left( \frac{300}{x} + 15 \right)^2$ . Thus, the function may be written as $(x+20)^2 + \left( \frac{300}{x} + 15 \right)^2$

We know that $x^2 + y^2 \geq 0$ with equality only when $x=y=0$ . $x+20 = 0 \iff x = -20$ and because $\frac{300}{-20} + 15 = 0$ , $x = -20$ is the value we are looking for.

We have that $\frac{90000}{x^2} + \frac{9000}{x} + 625 + 40x + x^2 = \frac{(x + 20)^2 (x^2 + 225)}{x^2} \ge 0$ for all $x \neq 0$ . Equality occurs for $x = -20$ .

$\begin{aligned} f(x) & = \frac {9000}{x^2} + \frac {9000}x + 625 + 40x + x^2 \\ & = 900\left(\frac {100}{x^2} + \frac {10}x + \frac 14\right) - 225 + 625 + (x^2 + 40x + 400) - 400 \\ & = 30^2 \left(\frac {10}x + \frac 12\right)^2 + (x+20)^2 \end{aligned}$

We note that $f(x) \ge 0$ . Using Titu's lemma , we have:

$\begin{aligned} f(x) & = 30^2 \left(\frac {10}x + \frac 12\right)^2 + (x+20)^2 \ge \frac {\left(30\left(\frac {10}x + \frac 12\right) + x+20\right)^2}{1+1} = \frac {(x^2 + 35x-300)^2}{2x^2} \end{aligned}$

Equality occurs when

$\begin{aligned} 30\left(\frac {10}x + \frac 12\right) & = x+20 \\ \implies x^2 + 5x - 300 & = 0 \\ (x+20)(x-15) & = 0 \\ \implies x & = \boxed{-20} & \small \color{#3D99F6} \text{when minimum }f(x) = 0 \text{ occurs.} \end{aligned}$