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Is there any way to solve this inequality? I asked my friend for help, but he couldn't do it. I can't use even derivatives and his solution was including them. So, after many transformations i have to show this inequality :

$$\frac{\ln x}{t} + \ln t > \ln \ln \ x$$ for $t > 1$

Is there a way to show it's true ? ( Wolfram says so... ). But if i over-complicated things, here's the original inequality : $$t \ln x < x^{1/t}$$ I have to show it's true for some large $x$. At my first inequality, it must be true for $x > 0$. If anyone would be interested in those weird transformations, i'll provide them.

I've spent over 3 hours in one go thinking about it, but i always fail...

Would be great to receive some cool hints or answer to this task.

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  • $\begingroup$ Your original inequality is equivalent to $x>t^t (\log x)^t$ and to show that this is true for $x$ big enough it is enough to show that $\lim_{x\to +\infty} \frac{x}{(\log x)^t}=+\infty$ for each $t>1$. $\endgroup$
    – Daniele A
    Jan 16, 2014 at 0:38
  • $\begingroup$ Thanks for your answer, but i'm not convinced about your solution. Why did you miss $t^t$? $\endgroup$ Jan 16, 2014 at 0:49
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    $\begingroup$ Because it is constant: the fact that $\lim_{x\to +\infty} \frac{x}{(\log x)^t} = +\infty$ means that for every $M>0$ we have that $x>M(\log x)^t$ for $x$ big enough. Taking $M=t^t$ gives the result that you want. $\endgroup$
    – Daniele A
    Jan 16, 2014 at 0:53
  • $\begingroup$ I couldn't prove it by induction that $x > (log x)^t$. How to show that this lim goes to infinity? Excluding hand waving :-P $\endgroup$ Jan 16, 2014 at 1:10
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    $\begingroup$ It is enough to prove this for $t=n\in\mathbb{N}$ because for every $t$ we can choose an integer $t>n$. Now, with the substitution $y=\log x$, it is enough to prove that $\lim_{y\to +\infty}\frac{e^y}{y^n}=+\infty$ for every $n\in\mathbb{N}$. To prove this, you could use the formula $e^y=\sum_{m=0}^{+\infty}\frac{y^m}{m!}$ that tells you in particular that $e^y\geq \frac{y^{n+1}}{(n+1)!}$ $\endgroup$
    – Daniele A
    Jan 16, 2014 at 1:24

1 Answer 1

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Find the minimum in $t$ of $\frac{\log(x)}{t}+\log(t)$. Its derivative is $0$ when $t=\log(x)$. Thus, $$ \begin{align} \frac{\log(x)}{t}+\log(t) &\ge\frac{\log(x)}{\log(x)}+\log(\log(x))\\ &=1+\log(\log(x))\\[4pt] &\gt\log(\log(x)) \end{align} $$


Without derivatives:

We have the inequality for all $y$: $1+y\le e^y$. If we let $y=-\log(u)$, we get $$ 1-\log(u)\le\frac1u $$ Consider $t=u\log(x)$. Then we have $$ \begin{align} \frac{\log(x)}{t}+\log(t) &=\frac1u+\log(u)+\log(\log(x))\\ &\ge1+\log(\log(x)) \end{align} $$

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  • $\begingroup$ It's very smart! Is there a way to not use derivatives? Most of us on studies know them already, but we are not officialy allowed to use them, stupid as it sounds, but maybe there is a way to do it? $\endgroup$ Jan 16, 2014 at 0:47
  • $\begingroup$ @Chris: I've added a proof without derivatives, but it does require the inequality $1+y\le e^y$, which can be proven with Bernoulli's Inequality, which can be proven by induction. $\endgroup$
    – robjohn
    Jan 16, 2014 at 1:17
  • $\begingroup$ Hey! Thanks for your edit. I got a question about a proof. Is it okay if i say that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \dots$ so, it's obviously greater than $1+x$. Is it enough to prove this? $\endgroup$ Jan 16, 2014 at 1:29
  • $\begingroup$ @Chris: for $x\ge0$, that suffices. However, note that you are using Taylor Series... $\endgroup$
    – robjohn
    Jan 16, 2014 at 1:30
  • $\begingroup$ Holy cow, i checked and i can't use Taylor series either. Those studies are going to give me a huge headache. I think i gonna sleep now and work on it tommorow. Thanks for your help ! $\endgroup$ Jan 16, 2014 at 1:44

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