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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.

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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
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@AndréNicolas He actually wants $x\geq 1$. –  Pedro Tamaroff Sep 27 '12 at 15:33
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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. –  TheClock Sep 27 '12 at 15:34
    
Okay, but now I think it's obivous that this is bigger than 0? –  TheClock Sep 27 '12 at 15:41
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4 Answers

up vote 3 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

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

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Sorry, but I meant x>=0. –  TheClock Sep 27 '12 at 15:51
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@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. :) –  TheClock Sep 27 '12 at 16:03
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@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. :=) –  TheClock Sep 27 '12 at 16:12
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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.

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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
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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 $x\geq1$.

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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.

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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? –  TheClock 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
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