Prove inequality $2e^x>x^3+x^2$

Question

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 $$f(x)>0 \text{ for } x \in \Bbb R?$$

Answer 1

• 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 $$ f(1+\sqrt{3}) = 1+\sqrt{3} - 2\log(2+\sqrt{3}) \approx 0.098135 > 0. $$

Answer 2

If x<-1, that's trivial. If x is between -1 and 0, it's easy to see that $2e^x>2e^{-1}>\max\{x^3+x^2\}.$

If x is between 0 and 1, it's easy to see that $2e^x>2>\max\{x^3+x^2\}.$

Let $c_1>1$ be the first root of $x^3+x^2=2e^1.$ Then between 1 and $c_1$, you have $2e^x>2e >\max\{x^3+x^2\}. $

I guess we can continue this process to get a proof.

Answer 3

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 $x>0$ 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 $e^x > \frac{1}{2} (x^3 + x^2)$ = $\sum_{n=1}^{\infty} \frac{x^n}{n!} > \frac{1}{2} (x^3 + x^2)$

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!