Show that $1/x > \log(x)^2$, for all real $x < 1$

Question

so i'm taking a Calc 1 class right now, and I am supposed to evaluate, whether there is a constant continuation of $f(x) = x^x$, where x approaches zero and is a positive real number.

I have a function $a(x) := e^x$, which is always higher than $f(x)$ for $x < e$ and $a(0) = 1$.

My other function is $b(x) := - \sqrt x +1$. I know, that: $b(0) = 1$.

Right now, I want to prove, that: $a(x) > b(x), \forall x \in (0, 1]$.

I already used the Bernoulli-Inequality, to get:

$-\sqrt x < \log(x^x) = x \log(x)$

Which should be equivalent to showing:

$1/x > \log(x)^2$

I already found out, that $\log(x)^2 = \log(1 / x)^2$ for $x < 1$, but still, im a bit stuck here and don't know how to proceed. Furhermore, I mustn't use differentiation, as we didn't introduce it yet in our course.

Any help would be appreciated, thanks!

Answer 1

For $x\le0$, the inequality makes no sense .

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For $x\in]0,1]$, set $y=\frac{1}{\sqrt x}$. Then $y\in[1,\infty[$. So $$\frac1x>\ln(x)^2\iff y^2>\ln\left(\frac1{y^2}\right)^2\iff y>-\ln\left(\frac1{y^2}\right)$$

The RHS is equal to $2\ln(y)$, so we have to prove $y>2\ln(y)$ for $y\in[1,\infty[$. Now,

$$y>2\ln(y)\iff \exp\left(\frac y2\right)>y,$$ which is true thanks to the classical inequality that for every $r\geq0$, $\exp(r)\geq 1+r+\frac{r^2}2$.

Indeed, setting $r=\frac y2$ gives us $$\exp\left(\frac y2\right)\geq\frac18 (y^2+4y+8)=\frac18 ((y-2)^2+8y+4)> y.$$

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Remark. In fact, this is harder to prove, but the inequality is true for all $$x\in\left]0, \exp(2 W(1/2))\right[,$$ where $W(1/2)\approx 0.35$ is the Lambert-W-function applied to $\frac12$.