<s>[INST] To prove an Inequality: $ ( x^2 +2x)e^x + (x^2-2 x)e^{-x} \ge 0$ 
$ \left(x^2  +2x\right)e^x + \left(x^2-2 x\right)e^{-x} \ge 0$.

I used photomath to plot its graph: $y=(x^{2}+2x))e^{x} + \frac{{x}^{2}-2x}{{e}^{x}}$

But how do I prove it without an image? Should I take the derivative of it and reason, please tell me the solution.
 [/INST] Rearranging gives
$$2x^2\frac{(e^x+e^{-x})}{2}+4x\frac{e^x-e^{-x}}{2}=2f(x)+4g(x)$$
Now we will use Taylor expansion !!!
note $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
$-e^{-x}=-1+x-\frac{x^2}{2!}+\frac{x^3}{3!}-....$
So,
$$\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+...$$
So $$x\frac{(e^x-e^{-x})}{2}=x^2+\frac{x^4}{3!}+\frac{x^6}{5!}+...\geq 0$$
So, $$4g(x)\geq 0$$(#)
Now by AM-GM we have $$2f(x)\geq2x^2\geq 0$$ (@)
By adding (@) and (#) gives the result!!
</s>