Prove inequality $2e^x>x^3+x^2$ If $x \in \Bbb R$, show that
$$2e^x>x^3+x^2$$
This inequality is right, see

Own ideas: If $x\in \Bbb R$,
$$f(x)=2e^x-x^3-x^2$$
$$f'(x)=2e^x-3x^2-2x$$
$$f''(x)=2e^x-6x-2$$
$$f'''(x)=2e^x-6$$
$$f''''(x)=2e^x>0$$
Like the symbol cannot judge $f$ sign.
So how can we show this $$f(x)>0 \text{ for } x \in \Bbb R?$$

*

*For $x \leq 1$, notice that
$$ x^3 + x^2 \leq x + 1 \leq e^x \leq 2e^x. $$

*For $x \geq 1$, it suffices to prove that $f(x) := \log(2e^x) - \log(x^3 + x^2) > 0$. Differentiating twice, 
$$ f'(x) = 1 - \frac{2}{x} - \frac{1}{x+1}, \qquad f''(x) = \frac{2}{x^2} + \frac{1}{(x+1)^2} $$
This shows that $f$ is strictly convex and attains global minimum on $[1, \infty)$ at $x = 1+\sqrt{3}$. Now the conclusion follows from
$$ f(1+\sqrt{3}) = 1+\sqrt{3} - 2\log(2+\sqrt{3}) \approx 0.098135 > 0. $$
