Prove $\forall x > 0$ that $(e^x - 1) \ln(1 + x) > x^2$

Question

Prove that $(e^x - 1) \ln(1 + x) > x^2$ for $x > 0$.

It is easy to prove that $e^x > 1 + x + \frac{x^2}{2}$ for $x > 0$. Thus, $e^x - 1 > \frac{x^2}{2}$ for $x > 0$.

Therefore $(e^x - 1) \ln(1 + x) > \frac{x^2}{2} \ln(1 + x) > x^2$.

$\frac{x^2}{2} \ln(1 + x) > x^2 \Rightarrow \ln(1+x) > 2 \Rightarrow x > e^2 - 1$.

How would I go about proving inequality for $x \in (0; e^2 - 1]$?

Answer 1

By the Cauchy-Schwarz inequality:

$$ \int_{0}^{x} e^{u}\,du \int_{0}^{x} \frac{1}{1+u}\,du \stackrel{\text{CS}}{\geq}\left(\int_{0}^{x}\sqrt{\frac{e^u}{1+u}}\,du\right)^2 $$ and since $e^u> 1+u$ by convexity, the RHS is $> \left(\int_{0}^{x}1\,du\right)^2 = x^2$.
We actually have $\frac{e^u}{1+u}\geq 1+\frac{u^2}{3}$ for any $u\geq 0$, hence the original inequality can be largely improved. Anyway, it is a very loose inequality for any large $x>0$.

Answer 2

Jack D'Aurizio’s answer is correct but terse, and might be hard to read for those who are not so good at mathematics. Diger provided a very different proof as an answer to a duplicate question, though on further reflection that proof might also be a bit hard to read.

Note that both this question and the duplicate question present statements as true when they are actually new statements to be proved. For example, this question says:

\begin{equation} \frac{x^2}{2} \ln(1 + x) > x^2. \tag{1} \end{equation}

But, at this point, the question has not yet established this to be true. It is just a new statement to be proved, and the question should make this clear.

Actually, it is even worse: the question never establishes that this is true for any value of $x$. Luckily, in this case (though not in all cases), it is valid to replace the implications with equivalences, and this would establish that equation $(1)$ is true for $x > e^2 – 1$.

Aside from this, this question’s proof is valid.

Note that Diger’s proof, at one point, says:

since $x> 0$, it suffices to check

\begin{equation} \log(1+x) - \frac{x}{1+x/2} = \log(1+x) - \left(2 - \frac{4}{x+2}\right) \geq 0\,. \end{equation}

The equality is already known to be true, but the inequality is a new statement to be proved; this is made clear by the phrase “it suffices to check”.