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



Own ideas: If $x\in \Bbb R$,

$$f(x)=2e^x-x^3-x^2$$
$$f'(x)=2e^x-3x^2-2x$$
$$f''(x)=2e^x-6x-2$$
$$f'''(x)=2e^x-6$$
$$f''''(x)=2e^x&gt;0$$

Like the symbol cannot judge $f$ sign.

So how can we show this $$f(x)&gt;0 \text{ for } x \in \Bbb R?$$
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 $f(x) := \log(2e^x) - \log(x^3 + x^2) &gt; 0$. 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
$$ f(1+\sqrt{3}) = 1+\sqrt{3} - 2\log(2+\sqrt{3}) \approx 0.098135 &gt; 0. $$
Consider that for all x in $\mathbb R$, $e^{x}$ > 0, so  $2e^{x}$ > 0 for all $\mathbb R$.So clearly for $x\leq 0$ in $\mathbb R$, the statement is true since $x^2 \geq 0$ and  $x^3 \leq 0$.  Since $|x^3| \geq x^2$ for x in $\mathbb R$, $x^3 + x^2 \leq 0$ for $x\leq 0$ in $\mathbb R$. 

Now consider where $x&gt;0$ in $\mathbb R$. Let's use the exponential power series since we know it converges for all real numbers. Assuming x > 0,rearranging and substituting gives $e^x &gt; \frac{1}{2} (x^3 + x^2)$ =  $\sum_{n=1}^{\infty} \frac{x^n}{n!} &gt; \frac{1}{2} (x^3 + x^2)$

The question now is which side grows larger as $x\rightarrow +\infty$. Clearly both go to infinity in this case. So consider the following:  

$\frac{e^x} {\frac{1}{2} (x^3 + x^2)}$  

Now use L'Hospital's rule: 

$\frac{e^x} {\frac{3}{2} (x^2 + x)}$ 

Apply again: 

$\frac{e^x} {3x + 1}$

One last time. 

$\frac{e^x} {3}$

Clearly as $x\rightarrow +\infty$ then  e^x > 3. That completes the proof. Whew!