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I need to compare $1-\frac{2}{3}\cdot3^{-\frac{2}{3}}\cdot \log_e9$ and $0$ without any computer

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  • $\begingroup$ You just need to see if it's positive or not. After some fiddling, you'll see that this is equivalent to seeing if $\displaystyle\log_e9 < \frac{3^\frac{5}{3}}{2}$. $\endgroup$ – user3482749 Nov 15 '18 at 12:02
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With the use of $\;9<e^3$ we get $$\frac{2}{3}\cdot3^{-\frac{2}{3}}\cdot \ln 9<\frac{2}{3}\cdot3^{-\frac{2}{3}}\cdot \ln e^3=\frac{2}{9^{1\over 3}}<\frac{2}{8^{1\over 3}}=1,$$ thus $$1-\frac{2}{3}\cdot3^{-\frac{2}{3}}\cdot \ln 9>0.$$

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It's easy to take conservative estimates $3^{-\frac{2}{3}}<\frac{1}{2}$, $\log_e9<\frac{5}{2}$ and conclude that the expression is positive.

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  • $\begingroup$ And why is $\log_e 3$ less than $\frac{5}{2}$? $\endgroup$ – Mark Tiukov Nov 15 '18 at 12:22
  • $\begingroup$ @МаркТюков: because $e\approx 2.7$, $e^2\approx 7.3$, $\sqrt{e} > 1.5$ $\endgroup$ – Vasya Nov 15 '18 at 12:41

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