0

What is the easiest way to show that for small enough values of $x \in \mathbb{R}$

$x(\log(x))^2 > \sqrt{x}$

i.e. there is some interval $[0,\lambda]$ on which the inequality is true. One idea is to observe that the two sides are equal at $x=0$ then show that the derivative of the left hand side is greater than the derivative of the right hand side for some interval $[0,\lambda]$, but that just shifts the problem to proving the inequality for the derivatives which is just as difficult. All my attempts seem to bring me to Lambert W function which I am not supposed to use.

Sine of the Time
  • 4,931
Noppawee Apichonpongpan
  • 986
  • 3
    This is false. $\lim_{x\to0^+}\sqrt x\log^2(x)=0.$ https://math.stackexchange.com/a/1975905 – Anne Bauval Dec 10 '23 at 13:04
  • Perhaps the question should be rephrased though, because it will hold on an interval $]\varepsilon,\lambda[$, not $[0,\lambda]$. – Daniel P Dec 10 '23 at 13:18
  • @DanielP But the question is for "small enough" values. Obviously for large values of $x$ the inequality will hold. – Mark Dec 10 '23 at 13:18
  • @DanielP It is false that "there is some interval $[0,\lambda]$ on which the inequality is true". – Anne Bauval Dec 10 '23 at 13:18
  • After some binary search, it seems to hold on the interval approximately $$]0.0001817,0.2394[$$ (assuming $\log = \ln$) – Daniel P Dec 10 '23 at 13:25
  • @DanielP also above $2.26043$. These are $e^{4 W_{-1}(-1/4)}$, $e^{4 W_{0}(-1/4)}$ and $e^{4 W_{0}(+1/4)}$ – Henry Dec 10 '23 at 13:33

1 Answers1

0

Here are three charts of $\sqrt{x}$ in black and $x(\log(x))^2$ in red over different ranges of $x$. You can see that for $x$ very small and positive, you have $x(\log(x))^2 < \sqrt{x}$, contrary to the assertion.

Looking at $x=e^{-n}$, you are comparing $n^2e^{-n}$ and $e^{-n/2}$, equivalent to comparing $n^4$ and $e^n$ and it should be easy to see that for $n \ge 9$ you have $n^4 < e^n$. This means that for $0 < x <e^{-9}\approx 0.0001234$ you have $x(\log(x))^2 < \sqrt{x}$.

You can improve that $e^{-9}\approx 0.0001234$ bound to about $0.000181697$ either numerically or using the Lambert W function as in $x = e^{4 W_{-1}(-1/4)}$, but you do not need to.

\sqrt{x} in black and x(\log(x))^2 in red

\sqrt{x} in black and x(\log(x))^2 in red

\sqrt{x} in black and x(\log(x))^2 in red

Henry
  • 157,058