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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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    Then you might note $\ln(ex)\ge\ln(e)=1$; so, $x\ln(ex)-\sqrt x\ge x-\sqrt x=\sqrt x(\sqrt x-1)\ge 0$. Or, better, look at Peter's answer.2012-09-27

4 Answers 4

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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.

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

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

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    Yeah, I just wanted you to know. :=)2012-09-27
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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 $x\ge 0$.2012-09-27
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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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    @JulianAssange See $m$y a$n$swer...2012-09-27