Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I continue to prove this?

Show that $$ x \ln(ex) - \sqrt{x}\geq 0 $$ for all $$ x \geq1 $$

My try:

$$\begin{eqnarray*} \\ \ln(e^x) + \ln(x^x) &\geq& \sqrt{x} \\ \\ \ln(e^x) &\geq& \sqrt{x} - \ln(x^x) \\ \\ e^x &\geq & e^\sqrt{x} e^{-\ln(x^x)} \\ \\ e^x&\geq& e^\sqrt{x} (1/x^x) \\ \\ e^x - \frac{e^\sqrt{x}}{x^x}&\geq& 0 \end{eqnarray*}$$

I "see" that this is bigger than 0, but I think that there are more calculations to do.

share|cite|improve this question
Let $x$ be positive but less than $1/e$. Then our expression is clearly negative. So you will have to change the condition $x\ge 0$ to something else. – André Nicolas Sep 27 '12 at 15:32
@AndréNicolas He actually wants $x\geq 1$. – Pedro Tamaroff Sep 27 '12 at 15:33
You have an error: $e^{-\ln x^x}\ne -x^x$; but rather $e^{-\ln x^x}= e^{\ln (1/x^x)}=1/x^x$. – David Mitra Sep 27 '12 at 15:33
@DavidMitra: Thanks. – Curtain Sep 27 '12 at 15:34
Okay, but now I think it's obivous that this is bigger than 0? – Curtain Sep 27 '12 at 15:41
up vote 4 down vote accepted

Consider the function

$$f\left( x \right) = x\log \left( {ex} \right) - \sqrt x = x\log x + x - \sqrt x $$

$\log x$ is positive for $x>1$ and negative for $0<x<1$.

And $x>\sqrt x$ for $x>1$, and $x<\sqrt x$ for $0<x<1$. Thus

$$\begin{cases} f(x)>0 \text{ for } x>1\\ f(x)<0 \text{ for } 0<x<1\\f(x)=0 \text{ for }x=1\end{cases}$$

share|cite|improve this answer
Sorry, but I meant x>=0. – Curtain Sep 27 '12 at 15:51
@JulianAssange I'm showing you the inequality is not true for $x\geq 0$. – Pedro Tamaroff Sep 27 '12 at 16:00
Sorry, I meant x >= 1. :) – Curtain Sep 27 '12 at 16:03
@JulianAssange I showed you that is true, too. Did you read the answer carefully? – Pedro Tamaroff Sep 27 '12 at 16:10
Yeah, I just wanted you to know. :=) – Curtain Sep 27 '12 at 16:12

Substitute 1/e for x. Of course 1/e is greater than zero. But ln(e*1/e) equals zero, so your Left Hand Side is negative, equals to -1/sqrt{e}. Therefore, the inequality you are trying to prove is false.

share|cite|improve this answer
You should probably edit this to say that it was for the original version of the problem, which claimed the inequality for all $x\ge 0$. – Brian M. Scott Sep 27 '12 at 20:32

Let $f(x)=x\ln(ex)-\sqrt x$. Note that $f'(x)=\ln(ex)+1 -\tfrac{1}{2\sqrt x}>0 \;(*)$ for $x\geq1$, and so $f$ is increasing from $x=1$. But $f(1)=0$, and so $f(x)\geq 0$ for $x\geq 1$.

$(*)$ This can be seen since $\ln(ex)+1$ is increasing, and $\tfrac{1}{2\sqrt x}$ is decreasing, so $f'(x)$ must be increasing. Since $f'(1)>0$, we thus must have this inequality for all $x\geq1$.

share|cite|improve this answer

The inequality is true for $x \geq 1$. Put $f(x)=x\ln({\rm e}x)-\sqrt{x}$ and use the derivative test to prove that.

share|cite|improve this answer
How does one prove that? I know that the function is acting like an exponential function, but is it enough to derivate and see that the interval from x >= 1 is growing? – Curtain Sep 27 '12 at 16:05
@JulianAssange: Just look at any calculus book under applications of derivative (finding maximum and minimum of functions). I'll try to post it later. – Mhenni Benghorbal Sep 27 '12 at 16:12
@JulianAssange See my answer... – SL2 Sep 27 '12 at 16:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.