I am not getting the nerve how to even start. Any hint will be appreciated.
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1Hint: $\pi^{1/\pi} < e^{1/e}$. Since both numbers are positive, we can exponentiate both sides by the same (positive) number and the inequality stays the same. – 2012ssohn Jun 04 '15 at 00:54
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The answer by Yuval Filmus there. ^ – Jun 04 '15 at 01:59
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Can you show that $x^{\frac{1}{x}}$ is maximized when $x = e$?
Then, that means that $\pi^{\frac{1}{\pi}} < e^{\frac{1}{e}} \dots$.
Michael Biro
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Hi sir, I have completed increasing & decreasing function but haven't started maxima & minima: so can you show me the proof? – Jun 04 '15 at 00:55
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Let $f(x) = \dfrac{\ln x}{x}, x > 1 \Rightarrow f'(x) = \dfrac{1-\ln x}{x^2} = 0 \iff x = e, f'(x) > 0 \iff x < e, f'(x) < 0 \iff x > e$. This means $x = e$ is a maxima of $f$. Since $\pi > 3 > e>1$, $f(e) > f(\pi) \Rightarrow \dfrac{\ln e}{e} > \dfrac{\ln \pi}{\pi} \Rightarrow \pi\ln e > e\ln \pi \Rightarrow \ln(e^{\pi}) > \ln(\pi^{e})\Rightarrow e^{\pi} > \pi^{e}$.
DeepSea
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Alternative proof that $\pi^e<e^\pi$, using the fact that $e^x\ge x+1$ for every value of $x$ (with equality iff $x=0$). Look at a graph to see why this inequality makes sense.
Proof: Let $x=\frac\pi e-1$ in the above inequality: \begin{align} e^{\pi/e-1}&>\frac\pi e\\ e^{\pi/e}&>\pi\\ e^\pi&>\pi^e \end{align}
Akiva Weinberger
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If you replace $\pi$ by $x$, this shows that $e^x > x^e$ if $x > e$. – marty cohen Jun 04 '15 at 03:05
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@martycohen Indeed. Though, you probably meant $x\ne e$ (because at $x=e$ it's an equality rather than an inequality); no reason to restrict $x>e$. (You'll want $x>0$, though, so that $x^e$ is always defined.) – Akiva Weinberger Jun 04 '15 at 03:07
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True, because what is needed is $e^x > 1+x$ which is true for $x \ne 0$. – marty cohen Jun 04 '15 at 03:12
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Hint: your inequality to the $\frac{1}{e\pi}$ power.
Hint 2: $\pi \approx 3.141$ and $e \approx 2.718$
MichaelChirico
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