<s>[INST] Proving an inequality is true (precalculus) How do I prove that
$(|x|+2)(|x^2+9|)-9|x^2-2| \ge 0$?
I tried using properties of absolute values such as triangle inequalities but so far I've got no luck.
The actual question was to prove that
$|\frac{x^2-2}{x^2+9}| \le \frac{|x|+2}{9}$. I tried using triangle inequality properties but only to the point where I get $|\frac{x^2-2}{x^2+9}| \le \frac{|x^2|+2}{9}$ which is different from what I wanted to prove.
 [/INST] Using $f(x) = (|x|+2)(x^2+9)-9|x^2-2|$ as defined in the other answer and the observation that $f(x) = f(-x)$, we prove it when $x \ge 0$ using a slightly different way: completing squares.
For $0 \le x \le \sqrt 2$, we have:
$$f(x) = (x+2)(x^2+9)-9(2-x^2) = x^3+11x^2+9x\ge0 \text{ since }x \ge 0$$
For $x \ge \sqrt 2$ we have:
$$\begin{align}f(x) &= (x+2)(x^2+9)-9(x^2-2) \\&= x^3-7x^2+9x+36\\&= x^3-7x^2+\frac{49}4x-\frac{13}4x+36\\&=x(x-3.5)^2+\frac14(144-13x)\end{align}$$
which is positive when $\sqrt 2 \le x \le 10$.
We also have:
$$f(x) = x^3-7x^2+9x+36 > x^2(x-7)$$
which is positive when $x \ge 7$.
This proves the result.
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