Solve $n < e^{6 sqrt{n}}$

Mathematics Asked on November 24, 2020

Find for which values of $$n in mathbb{N}$$ it holds that $$n < e^{6 sqrt{n}}.$$

I tried to use the inequality $$(1 + x) leq e^x$$, but from this, I can only find that the inequality holds for $$n > 36$$. But I need to get $$n$$ as small as possible.

I also tried the induction on $$n$$, but I stucked in the induction step. In particular, in showing that $$e^{6sqrt{n}} + 1 leq e^{6sqrt{n+1}}$$.

I appreciate any help and suggestions.

We have $$e^{3x} > 3x ge x$$ for $$xge 0$$. Apply this to $$x = sqrt{n}$$ to get $$e^{3sqrt{n}} > sqrt{n}$$. Square both sides and you have $$e^{6sqrt{n}} > n$$ for all $$nge 0$$.

Correct answer by jjagmath on November 24, 2020

The inequality holds for all positive $$x$$, so effectively that a proof seems overkill :-)

(Notice that $$0 and $$1.)

Or using Taylor,

$$n<1+6sqrt n+18n+cdots$$

Answered by Yves Daoust on November 24, 2020

(Based on hint from user2661923, see the comments above.)

I set $$f(x) = e^{6sqrt{x}} - x$$. The first derivation of $$f(x)$$ is $$f'(x) = 3 frac{e^{6sqrt{x}}}{sqrt{x}} - 1.$$

We note that $$f(1) = e^6 - 1 > 0$$. Moreover, it is not hard to check that $$f'(x) > 0$$ for $$x in [1, +infty)$$. Thus $$f$$ is increasing on $$[1, +infty)$$ and $$f(1) > 0$$. Therefore $$f(x) > 0$$ for all $$x in [1, +infty)$$. It implies that $$x < e^{6sqrt{x}}.$$

And my question easily follows.

Answered by Kapur on November 24, 2020

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