Neat way to prove that $x^2\gt (1+x)(\ln(1+x))^{2}, ~\forall x\gt 0$

Question

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.

Answer 1

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 $$x^2\gt (1+x)\ln^2(1+x)$$

Answer 2

Denote $y=\ln(1+x)$, then

$$x^2 - (1+x)(\ln(1+x))^2 = (e^y -1)^2 - e^y y^2\\ = e^{2y}-2e^{-y}+1 - e^y y^2 = e^y (e^y + e^{-y}-2 - y^2)\\ = e^y\left(\sum_{n=0}^\infty \frac{2 y^{2n}}{(2n)!} - 2 - y^2 \right) = e^y\sum_{n=2}^\infty \frac{2 y^{2n}}{(2n)!} >0.\blacksquare $$

Answer 3

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.

Answer 4

let $x+1=e^t$ then $t>0$.

to prove $(1)$ is:we will prove $$F(t)=e^t+\frac{1}{e^t}-2-t^2>0$$ 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 $t>0$

Answer 5

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.