How to prove that $- x\ln{(x)}\le\sqrt{x}$ for $0

Question

I tried square both side (since both of them are positive) $ x^2\log^2{(x)}\le x$ , which implies $ x\log^2{(x)}\le 1$ but to no avail. I also tried solve it in exponential form, which would be $e^{(- x\log{(x)})} = x^{-x} \le e^{\sqrt{x}}$ and thus $1 \le e^{\sqrt{x}}x^{x}$. I feel like it is really close to the actual answer but don't know how to proceed from here. I already tried differentiating as well as suppose for contradiction that there exist an $x$ such that $1 > e^{\sqrt{x}}x^{x}$, but neither of them works: differentiating $e^{\sqrt{x}}x^{x}$ to show that it is strictly increasing would end up stuck at proving $- \sqrt{x}\ln{(x)}\le1$, which simply goes back to the beginning...

Answer 1

Let $\displaystyle f(x)=\frac{1}{\sqrt{x}}+\ln x$. Then $\displaystyle f'(x)=-\frac{1}{2(\sqrt{x})^3}+\frac{1}{x}=\frac{2\sqrt{x}-1}{2(\sqrt{x})^3}$.

So, $f'(x)>0$ for $\displaystyle \frac{1}{4}<x<1$ and $f'(x)<0$ for $\displaystyle 0<x<\frac{1}{4}$.

$f$ is continuous and hence $f$ attains its absolute minimum at $\displaystyle x=\frac{1}{4}$.

$$f\left(\frac{1}{4}\right)=2+\ln \frac{1}{4}=\ln\left(\frac{e^2}{4}\right)>\ln(1)=0$$

Answer 2

After dividing by $x$ and using a logarithm relation, the inequality is equivalent to $$\ln\left(\frac1x\right)\leq\sqrt{\frac1x},\quad0<x<1$$ which, after replacing $x$ by $1/x$ is equivalent to $$\ln\left(x\right)\leq\sqrt{x},\quad x>1$$ It is more well-known that $\ln(x)$ grows slower than $\sqrt{x}$. Do you have access to a resource that proves $\ln\left(x\right)\leq\sqrt{x}$ for $x>1$?

Answer 3

By expanding power series of $e^t$, one gets $1+t<e^t$ for $x>0$. By substituting $y=e^t$ and some calculation, one gets $\ln y<y-1$ as in my comment. Now, let $y=\sqrt x$ to get$$\ln x<2(\sqrt x-1)$$ By using this, we get$$-x\ln x<-2x(\sqrt x-1)$$ so it is enough to prove$$-2\sqrt x(\sqrt x-1)\le1$$ and by letting $u=\sqrt x$, this is equivalent to$$-2u^2+2u-1\le 0$$ for $0<u<1$. Can you finish this from here?

Answer 4

For $0<x<1$ we have

• $- x\ln{(x)}\le\sqrt{x} \Leftrightarrow -\ln x \leq \frac{1}{\sqrt x} \Leftrightarrow 0\leq \frac{1}{\sqrt x} + \ln x$
• consider $f(x) = \frac{1}{\sqrt x}+\ln x \Rightarrow f(1) >0$
• $f$ has on $(0,1)$ a local minimum at $x= \frac{1}{4}$ with $f(\frac{1}{4}) > 0$
• $\Rightarrow f(x) > 0$ on $(0,1)$

Answer 5

I did not see your requirement of 0 < x < 1. Oh well, the procedure might help another.

My math is a little rusty, and I am trying to get back into it. I think this problem can be solved with limits.

First, let us move the negative onto the other side and flip the inequality

$$ x \ln x \geq - \sqrt x $$

Now, let us take the exponential

$$ x^{x} \geq e^{- \sqrt x}$$

Now, taking the limits as x approaches 0 from the right side

$$ \lim_{x \to 0^+} x^{x} = 1$$ $$ \lim_{x \to 0^+} e^{- \sqrt x} = 1$$

This solves the "equal" part of the inequality. Next, we take the limit to infinity. Since x is under a square root sign, we must remain in the positive.

$$\lim_{x \to +\infty} x^x = + \infty$$ $$\lim_{x \to +\infty}e^{- \sqrt x} = 0$$

$$x^{x} \geq e^{- \sqrt x}$$

If you graph the functions, you see that the latter holds true.

References:

[1] Anton, Howard. (1992). Calculus with analytic geometry-4th Edition. Anton Textbooks, Inc.

Answer 6

Use that $$\log(x) < \sqrt{x} - \frac1{\sqrt{x}}$$ for $x>1$ (see here for example). Substitute $x\leftarrow x^{-1}$ and then multiply both sides by $x$ to find the stronger inequality $$-x\log(x) < \sqrt{x}(1-x)$$ for $0<x<1$.