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.

HINT:
$f(x) =(x^2 + 2 x) e^x$ has a Taylor series with all coefficients positive. So $f(x) + f(-x)$  will have a Taylor series with all coefficients $\ge 0$.
$\bf{Note}$: the function $g(x) = x e^x$   has a positive Taylor expansion at $0$, so $g(x) + g(-x) \ge 0$.  We get  stronger inequalities
$(x^2 +a x) e^x + (x^2 - a x) e^{-x}\ge 0$
by decreasing $a$. Now the Taylor expansion of the above is:
$$(x^2 + a x) e^x + (x^2 - a x) e^{-x}= 2(1+a) x^2 + \frac{3+a}{3}x^4 + \cdots + \frac{2(2n-1+ a)}{ (2n-1)!} x^{2n} + \cdots$$
For $a=-1$ we  get
$$(x^2 - x)e^x + (x^2 + x) e^{-x} \ge 0$$
