Neat way to prove that $x^2\gt (1+x)(\ln(1+x))^{2}, ~\forall x\gt 0$ I am supposed to prove the following inequality. One standard method that can be put to use is considering a function $f(x)=x^2-(1+x)(\ln(1+x))^2$ and take its derivative to comment on the behaviour of its increase and arrive at concluding the inequality using the fact that $f(0)=0$.

$$x^2\gt (1+x)(\ln(1+x))^2,~~ \forall x\gt 0$$

I was wondering if there was a neater way to come up with the proof, especially as this is a contest-practice problem. Any hints are appreciated. Thanks.
 A: Note
\begin{align}
\frac x{\sqrt{1+x}}-\ln(1+x)
&=\int_0^x \frac{t+2}{2(1+t)^{3/2}}dt-\int_0^x \frac1{1+t}dt\\
&=\int_0^x \frac{t+2-2\sqrt{1+t}}{2(1+t)^{3/2}}dt 
=\int_0^x \frac{(\sqrt{1+t}-1)^2}{2(1+t)^{3/2}}dt >0
\end{align}
Thus, $\frac x{\sqrt{1+x}}>\ln(1+x)$ and
$$x^2\gt (1+x)\ln^2(1+x)$$
A: First, take the square root on both sides, turning the inequality into $x\gt\sqrt{x+1}\ln(x+1)$. Next, let $u^2=x+1$ and rewrite the inequality to prove as
$${u^2-1\over u}\gt\ln(u^2)\quad\text{for }u\gt1$$
Finally, let $f(u)=u-{1\over u}-2\ln u$, noting that $f(1)=0$. We see that
$$f'(u)=1+{1\over u^2}-{2\over u}={u^2-2u+1\over u^2}=\left(u-1\over u \right)^2\gt0$$
for $u\gt1$ and thus $f(u)\gt0$ for $u\gt1$, which proves the desired inequality.
A: Denote $y=\ln(1+x)$, then
$$x^2 - (1+x)(\ln(1+x))^2 = (e^y -1)^2 - e^y y^2\\
= e^{2y}-2e^{-y}+1 - e^y y^2 = e^y (e^y + e^{-y}-2 - y^2)\\
= e^y\left(\sum_{n=0}^\infty \frac{2 y^{2n}}{(2n)!}  - 2 - y^2 \right) = e^y\sum_{n=2}^\infty \frac{2 y^{2n}}{(2n)!} >0.\blacksquare
$$
A: let $x+1=e^t$ then $t>0$.
to prove $(1)$ is:we will prove $$F(t)=e^t+\frac{1}{e^t}-2-x^2>0$$
notice that $$F''(t)\ge 0 \tag{$F'$ is increasing}$$ and $F''(t)=0$ only when $t=0$
$$F'(0)=F(0)=0$$
which means  $F(t)>0$ for $t>0$
A: Hint : It's equivalent to :
$$x+1-1\geq \sqrt{x+1}\ln(x+1)$$
Putting $x+1=y$ it becomes :
$$y-1\geq\sqrt{y}\ln(y)$$
Now we substitute : $y=e^u$ and introduce the function :
$$f(u)=e^u-1-e^{\frac{u}{2}}u$$
The derivative is :
$$f'(u)=e^u-e^{\frac{u}{2}}-\frac{1}{2}e^{\frac{u}{2}}u$$
Now factorize and use $e^x> x+1\quad \forall x>0$
Conclude.
