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This question already has an answer here:

How can we show that $$e^{-2\lambda t}\lambda^2\le\frac1{e^2t^2}\tag1$$ for all $\lambda,t\ge0$?

Applying $\ln$ to both sides yields that $(1)$ should be equivalent to $$t\lambda\le e^{t\lambda-1}\tag2.$$ So, if I did no mistake, it should suffice to show $x\le e^{x-1}$ for all $x\ge0$. How can we do this?

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marked as duplicate by Martin R, Community Mar 20 at 19:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The substitution $u=x-1$ transforms $x \le e^{x-1}$ into $u+1 \le e^u$, which is answered in the above-mentioned Q&A. $\endgroup$ – Martin R Mar 20 at 17:57
  • $\begingroup$ Note that applying the square root to (1) gives (2), not an application of the logarithm. $\endgroup$ – Martin R Mar 20 at 18:33
  • $\begingroup$ @MartinR Both is legitimate. I've applied the logarithm, rearranged and then the exponential again. $\endgroup$ – 0xbadf00d Mar 20 at 18:56
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The line $y=x$ is tangent on $y=e^{x-1}$ in $x=1$ and since $e^{x-1}$ is convex, then it lies upper that any tangent of it, specially $x\le e^{x-1}$.

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  • $\begingroup$ The question has been asked and answered many times. A possible duplicate target was already pointed out. (Your answer is essentially math.stackexchange.com/a/504671/42969.) $\endgroup$ – Martin R Mar 20 at 17:50
  • $\begingroup$ The question you mentioned in the comments is not a duplicate of this one unless if $e^x-1=e^{x-1}$. A duplication does not make an answer wrong but if I knew the duplication (though I could guess a bit) I wouldn't have answered it unless through comments... $\endgroup$ – Mostafa Ayaz Mar 20 at 17:53
  • $\begingroup$ The simple substitution $u=x-1$ transforms $x \le e^{x-1}$ into $u+1 \le e^u$ .... $\endgroup$ – Martin R Mar 20 at 17:55
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    $\begingroup$ Probably it would have been better spending my time to find a duplicate rather than saving that time for answering :) $\endgroup$ – Mostafa Ayaz Mar 20 at 17:58
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If i take the ln on both sides we get $$-2\lambda t +2\ln(\lambda)\le -2-2\ln(t)$$ and this is not the result given above!

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  • $\begingroup$ Yes, OP applied the square root, not the logarithm. But (1) is equivalent to (2). $\endgroup$ – Martin R Mar 20 at 18:48
  • $\begingroup$ Ok, i can read, he/she has written "Applying ln ln to both sides yields that (1) (1) should be equivalent to " $\endgroup$ – Dr. Sonnhard Graubner Mar 20 at 18:52
  • $\begingroup$ Simply divide by $2$, use $\ln(\lambda t)=\ln\lambda+\ln t$ and apply the exponential function again. Then you obtain $(2)$. $\endgroup$ – 0xbadf00d Mar 20 at 19:31

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