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I know $$\forall a\in(1,\infty), B\in(0,\infty), x\in(0,\infty)$$ $$a^x\geq \left(\frac{ex\ln(a)}{B}\right)^{B}$$ can be proved using AM-GM. Is there a simple way to show the inequality holds in all those cases? This comparison seems to me to not be equal anywhere so far as I have noticed, so if you replace $e$ with a larger number such that the comparison is equal, what is the maximum number to replace $e$?

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  • $\begingroup$ Since $x \ln(a) = a^x$ this can be simplified to $k \ge \left(\frac{e \ln(k)}{B}\right)^B$ for $k>1$, $B>0$. $\endgroup$ – Aeryk Oct 7 '14 at 20:54
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Fix $a^x = k$ and treat the right-hand side as a function $f(B) = \left(\frac{e \ln(k)}{B}\right)^B$. Then $$\lim_{B \to 0} f(B) = 1 \text{ and }\lim_{B\to\infty} f(B) = 0$$ and $$f'(B) = e^B \left(\frac{\ln (k)}{B}\right)^B \ln\left(\frac{\ln (k)}{B}\right).$$ So the only critical point is at $B=\ln(k)$. One can check that the second derivative is negative and so it is a local maximum. Then note that $f(\ln(k)) = k$.

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  • $\begingroup$ Your limits should read $$\lim_{B \to 0} f(B) = 1$$ and $$\lim_{B\to\infty} f(B) = 0$$ . Also, the critical point, $B=\ln(k)$, is the global maximum of $f(B)$, not the local minimum. $\endgroup$ – Nazgand Oct 8 '14 at 5:08
  • $\begingroup$ So true. This is what happens in the haste of trying to get out of the office at the end of the day. Corrected now, though the idea remains the same. $\endgroup$ – Aeryk Oct 8 '14 at 13:56
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$$a^x\geq \left(\frac{ex\ln(a)}{B}\right)^{B} \\ \iff x\ln a\geq B\ln \left(\frac{ex\ln a}{B}\right) \iff \frac{x\ln a}{B} \geq \ln \left(e\frac{x\ln a}{B}\right) = 1 + \ln\left(\frac{x\ln a}{B}\right) $$

Now it is the well known inequality $$ \ln(1+u)\le u $$with $$ u = \frac{x\ln a}{B} - 1 $$

There is equality iff $$ u = 0 \iff x\ln a = B $$

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The way I proved here this inequality is by using the Weighted Arithmetic Mean - Geometric Mean Inequality using a Taylor series as the weights for another Taylor series. The method used was outlines here.

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