I need to show that $x > -1 \implies x^2 \ge (1+x)(\log(1+x))^2$
I'm not sure how to go about this. I have thought about showing that the derivative of $x^2 - (1+x) \{\log(1+x)\}^2$ is positive for $x>0$ and negative for $x<0$ but that isn't much easier, and Taylor series doesn't really work either. Any hints?
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Let $y=x+1$ then
$$x > -1 \implies x^2 \ge (1+x) \{\log(1+x)\}^2$$
is equivalent to
$$y > 0 \implies (y-1)^2 \ge y \log^2 y$$
$$\iff -\frac{|y-1|}{\sqrt y}\le\log y\le\frac{|y-1|}{\sqrt y} \iff e^{-\frac{|y-1|}{\sqrt y}}\le y\le e^{\frac{|y-1|}{\sqrt y}}\iff 1\le ye^{\frac{|y-1|}{\sqrt y}} \le e^{2\frac{|y-1|}{\sqrt y}}$$
which is true, indeed for $1\le ye^{\frac{|y-1|}{\sqrt y}}$
$$ye^{\frac{|y-1|}{\sqrt y}}\ge y\left(1+\frac{|y-1|}{\sqrt y}\right)=y+y\frac{|y-1|}{\sqrt y}\ge 1$$
$$\iff y\sqrt y+y|y-1|\ge \sqrt y\iff y|y-1|\ge\sqrt y(1-y)$$
and for $ye^{\frac{|y-1|}{\sqrt y}} \le e^{2\frac{|y-1|}{\sqrt y}} \iff e^{\frac{|y-1|}{\sqrt y}}\ge y $ note that
$$\iff e^{\frac{y-1}{\sqrt y}}\ge 1+\frac{y-1}{\sqrt y}+\frac{(y-1)^2}{2y}+\frac{(y-1)^3}{6y\sqrt y}\ge y$$
$$\frac{y-1}{\sqrt y}+\frac{(y-1)^2}{2y}+\frac{(y-1)^3}{6y\sqrt y}\ge y-1$$
$$\frac{1}{\sqrt y}+\frac{(y-1)}{2y}+\frac{(y-1)^2}{6y\sqrt y}\ge 1$$
$$6y+3y\sqrt y-3\sqrt y +y^2-2y+1\ge 6y\sqrt y$$
$$y^2+4y+1\ge 3y\sqrt y$$
which is true indeed let $y=z^2>1$
$$y^2+4y+1\ge 3y\sqrt y\iff z^4-3z^3+4z^2+1\ge0$$
and
$$z^4-3z^3+4z^2+1\ge z^4-4z^3+4z^2+1=(z^2-2z)^2+1\ge0 \quad \square$$
On
HINT: write $$\left(\frac{x}{\sqrt{x+1}}\right)^2\geq \left(\log(x+1)\right)^2$$ and split into two cases $$\left(\frac{x}{\sqrt{x+1}}-\log(x+1)\right)\left(\frac{x}{\sqrt{x+1}}+\log(x+1)\right)\geq 0$$
On
It can be easier if you set $1+x=t$, so the inequality becomes $$ (t-1)^2\ge t(\log t)^2 $$ If we consider the function $$ f(t)=(t-1)^2-t(\log t)^2 $$ defined for $t>0$, we have $$ \lim_{t\to0}f(t)=1,\qquad \lim_{t\to\infty}f(t)=\infty $$ Moreover $$ f'(t)=2(t-1)-(\log t)^2-2\log t $$ and $$ f''(t)=2-2\frac{\log t}{t}-\frac{2}{t}=\frac{2}{t}(t-\log t-1) $$ It's easy to see that $f''(t)>0$ except for $t=1$, where $f''(1)=0$. Thus $f'$ is strictly increasing, hence only vanishes once.
As $f'(1)=0$, the function $f$ has an absolute minimum at $t=1$.
Alternative solution: the inequality is the same as $$ \lvert\log t\rvert\le\frac{\lvert t-1\rvert}{\sqrt{t}} $$ and we can study it for $t\ge1$; indeed, if $$ f(t)=\frac{t-1}{\sqrt{t}}-\log t=\sqrt{t}-\frac{1}{\sqrt{t}}-\log t $$ we have $$ f(1/t)=-\frac{t-1}{\sqrt{t}}+\log t=-f(t) $$ and so $|f(1/t)|=|f(t)|$.
Well, now it's easier to study $g(t)=f(t^2)$: $$ g(t)=t-\frac{1}{t}-2\log t $$ Since $$ g'(t)=1+\frac{1}{t^2}-\frac{2}{t}=\frac{t^2-2t+1}{t}\ge0 $$ the function $g$, and so also $f$, is increasing for $t\ge1$.
Hence $f$ has a minimum at $1$.
On
the derivative is $f'(x)=2x-2\log(1+x)-\log(1+x)^2$. By setting $t=\log(x+1)$ one gets $h(t)=2e^t-2t-t^2-2$. We get $h'(t)=2e^t-2t-2\geq 0$, by the Taylor expansion of $e^t$. But $h(0)=0$, so for $t<0$, which corresponds to $x<0$, we have $f'(x)<0$ and for $t>0$ we have $f'(x)>0$. Also $f(0)=0$, so $f\geq 0$.
On
Let $x=u^2-1$ with $u\gt0$. The inequality to prove becomes $(u^2-1)^2\ge u^2(\log(u^2))^2$, or
$$\left| u-{1\over u}\right|\ge2|\log u|$$
Since each side is now invariant under $u\to1/u$, it suffices to prove the inequality for $u\ge1$, in which case we can remove the absolute value signs and write the inequality to prove as
$$f(u)=u-{1\over u}-2\log u\ge0$$
for $u\ge1$. But this is now easy: We see that $f(1)=0$ and
$$f'(u)=1+{1\over u^2}-{2\over u}=\left(1-{1\over u}\right)^2\ge0$$
On
With the substitution $x = e^y - 1$ the inequality is equivalent to $$ e^y + e^{-y} \ge 2 + y^2 $$ for all $y \in \Bbb R$, and that is true because $$ \begin{align} e^y + e^{-y} &= \left(1+ \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + \frac{y^4}{4!}+ \cdots \right) \\ & \quad +\left(1- \frac{y}{1!} + \frac{y^2}{2!} - \frac{y^3}{3!} + \frac{y^4}{4!}- \cdots \right) \\ &= 2\left( 1+ \frac{y^2}{2!}+ \frac{y^4}{4!} + \cdots\right) \\ &\ge 2 + y^2 \ . \end{align} $$
Let $y = x+1$
In general, $y^y = e^{y\ln y} \implies e^y <= y^y <= e^{y\ln y} <= e^{y^2}$
Hence, $y^{y\ln y} <= e^{y(\ln y)^2} <= e^{y^2 \ln y}$
we need to prove that $e^{y(\ln y)^2} <= e^{y^2}$ which can be trivially proven.