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.
 A: Another proof without derivative:
Note that it is enough to prove the inequality for $x\geq 0$ because if x is negative then putting $y=-x$ we get $LHS=(y^2-2y)e^{-y}+(y^2+2y)e^y$ which is the same as before.So in other words the LHS is an even function.Let $x\geq 0$ so
$$(x^2+2x)e^x+(x^2-2x)e^{-x}=x^2(e^x+e^{-x})+2x(e^x-e^{-x})=x^2(e^x+e^{-x})+2x\frac{(e^{2x}-1)}{e^x}$$
This is greater than or equal to 0 for non negative x
A: Make a Variation chart :

*

*Compute the derivative $f'$ of $f : x \mapsto \left(  { x  }^{ 2  }  +2x  \right)   { e  }^{ x  }  + \frac{  { x  }^{ 2  }  -2x  }{  { e  }^{ x  }    } $

*Show that $f'(x) \leqslant 0$ for $x < 0$ and $f'(x) \geqslant 0$ for $x > 0$.

Edit Instead of studying the too complex $f$, it is way easier to study the function $g : x \mapsto e^xf(x)$
. Indeed, for any $x$ we have $f(x) \geqslant 0 \Longleftrightarrow g(x) \geqslant 0$. This way, we only need to determine the variation chart of $g$, whith the nicer expression $g(x) = (x^2 + 2x) e^{2x} + x^2 - 2x$ .
A: Re-factor it, and use hyperbolic functions.
$$(x^2+2x)e^x+(x^2-2x)e^{-x}$$
$$=x^2(e^x+e^{-x})+2x(e^x-e^{-x})$$
$$=2x^2\cosh(x)+4x\sinh(x)$$
Since $\cosh(x)\geq1\geq0$ and $x^2\geq0$, the first term is non-negative.
Since $\sinh(x)$ has the same sign as $x$, the second term is non-negative.
Hence the sum is non-negative.
A: You can factor out $x$, and multiply by $e^x$
Then consider $(x+2)e^{2x} \ge x-2$
You see $x+2 \ge x-2$ because they have same slope and the left side has a y intercept of positive 2 while the other is negative 2. More precisely test any point and the. The slope shows the rate of change is always the same. Further you see the difference is always constant, if 4.
And then the function $f(c)=e^{2x}$ only increases that difference when x is greater then zero.
When it is less than zero, there is point at $x=-2$ where the left sides growth will no longer make it less then the right. Then there is the range from zero to $x=-2$ To show.
A: 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!!
A: 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
$$(x^2 - x)e^x + (x^2 + x) e^{-x} \ge 0$$
