3

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.

Xoque55
  • 4,384
Krzysztof Lewko
  • 2,312
  • 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 $t>1$. – Daniele A Jan 16 '14 at 00:38
  • Thanks for your answer, but i'm not convinced about your solution. Why did you miss $t^t$? – Krzysztof Lewko Jan 16 '14 at 00:49
  • 1
    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. – Daniele A Jan 16 '14 at 00:53
  • 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 – Krzysztof Lewko Jan 16 '14 at 01:10
  • 1
    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 $e^y\geq \frac{y^{n+1}}{(n+1)!}$ – Daniele A Jan 16 '14 at 01:24
  • I think you meant to write ${\ln x\over t}\lt x^{1/t}$ for the original inequality, not $t\ln x$ on the LHS. (As written, the original inequality is clearly false when $x\gt1$ for sufficiently large $t$.) One could also write the correct inequality as $\ln x\lt tx^{1/t}$. – Barry Cipra Jan 16 '14 at 01:24
  • @BarryCipra, original inequality is correct. Like i mentioned in my post, it doesn't have to be true for all $x$, but it has to be true for some large $x$ :-) – Krzysztof Lewko Jan 16 '14 at 01:33
  • @DanieleA, thank you for your tips, it's very clear now. – Krzysztof Lewko Jan 16 '14 at 01:45
  • @Chris: I'm glad that I could help. – Daniele A Jan 16 '14 at 11:44
  • @Chris, color me confused, then. I thought the two inequalities were supposed to be equivalent. – Barry Cipra Jan 16 '14 at 11:55
  • @BarryCipra, they are equivalent. It is: showing first is true will imply that second is true too, and vice versa. – Krzysztof Lewko Jan 16 '14 at 13:33
  • @Chris, they are not equivalent. When you take the log of the original inequality, the $\ln t$ goes with the $\ln\ln x$, not with the $\ln x\over t$. – Barry Cipra Jan 16 '14 at 15:20

1 Answers1

3

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 $$ \begin{align} \frac{\log(x)}{t}+\log(t) &=\frac1u+\log(u)+\log(\log(x))\\ &\ge1+\log(\log(x)) \end{align} $$

robjohn
  • 345,667
  • 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? – Krzysztof Lewko Jan 16 '14 at 00:47
  • @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. – robjohn Jan 16 '14 at 01:17
  • 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? – Krzysztof Lewko Jan 16 '14 at 01:29
  • @Chris: for $x\ge0$, that suffices. However, note that you are using Taylor Series... – robjohn Jan 16 '14 at 01:30
  • 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 ! – Krzysztof Lewko Jan 16 '14 at 01:44