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.
 A: 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.
A: If $f(x)=\bigl(|x|+2\bigr)(x^2+9)-9|x^2-2|$, then you always have $f(-x)=f(x)$. So, you only have to prove it when $x\geqslant0$, in which case $f(x)=(x+2)(x^2+9)-9|x^2-2|$.
If $x\in\left[0,\sqrt2\right]$, then $f(x)=x^3+11 x^2+9 x$. So, $f'(x)>0$ in this interval. Since $f(0)=0$, you know that $f(x)\geqslant0$ there.
And if $x\in\left[\sqrt2,\infty\right)$, then $f(x)=x^3-7 x^2+9 x+36$, which is decreasing when $x\in\left[\sqrt2,\frac13\left(7+\sqrt{22}\right)\right]$ and increasing when $x\in\left[\frac13\left(7+\sqrt{22}\right),\infty\right)$. But$$g\left(\frac13\left(7+\sqrt{22}\right)\right)=\frac{853-44\sqrt{22}}{27}>0.$$ So, $f(x)\geqslant0$ there too.
