Prove that $\,\sqrt [n] n < 1 + \sqrt{\frac{2}{n}}\,$

Question

I am having difficulty proving the following inequality: $$ \sqrt[n]{n} < 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n. $$ I am trying to use mathematical induction but I am having trouble going from the left side to the right side (in the induction step) and vice versa.

I also tried to take the power of $n$ on both sides and somehow use the binomial expansion but I am having difficulty with that. Thank you!

Induction would be difficult at best. Binomial expansion should yield the result quickly, what problems had you with that? — Daniel Fischer, Mar 13 '14 at 15:09
$n \in N$, it should be $n \geq 1$. I think it was more of the algebra and showing that it must be the case that the sum was greater than n — InsigMath, Mar 13 '14 at 15:11

Yiorgos S. Smyrlis · Accepted Answer · 2014-10-24T22:53:01.660

21

We shall show that $\,\Big(1+\sqrt{\frac{2}{n}}\,\Big)^n>n$.

For $n=1$ it is obvious.

Assume that $n\ge 2$, then according to the Binomial Formula: \begin{align} \left(1+\sqrt{\frac{2}{n}}\right)^n&=\binom{n}{0}+\binom{n}{1}\sqrt{\frac{2}{n}}+\binom{n}{2}\left(\sqrt{\frac{2}{n}}\right)^{\!2}+\cdots+\binom{n}{n}\left(\sqrt{\frac{2}{n}}\right)^{\!n} \\ &\ge 1+n\sqrt{\frac{2}{n}}+\frac{n(n-1)}{2}\cdot\frac{2}{n}=n+\sqrt{2n}>n. \end{align}

edited Oct 24 '14 at 22:53

answered Mar 13 '14 at 15:09

Yiorgos S. Smyrlis

83,933

How did you get any of that if you don't mind me asking (I mean the stuff right of the inequality) – InsigMath Mar 13 '14 at 15:12
@InsigMath ''according to the binomial theorem'' – OBDA Mar 13 '14 at 15:12
@InsigMath: The first thing that came to my mind was to show that $$\left(1+\sqrt{\frac{2}{n}}\right)^n>n.$$ And after that the only available tool is the binomial expansion. – Yiorgos S. Smyrlis Mar 13 '14 at 15:15
Yeah but the binomial expansion is a very large expression, you clearly did a lot of bounding at some point. – InsigMath Mar 13 '14 at 15:19
ohhh I think I see what's going on here! I might have to just try this with pencil and paper – InsigMath Mar 13 '14 at 15:20
2

Sorry I understand now! Thank you! – InsigMath Mar 13 '14 at 15:22
This is a direct proof, not an induction. – N. S. Oct 24 '14 at 22:48
That's right. Corrected now. – Yiorgos S. Smyrlis Oct 24 '14 at 22:53

score 0 · Answer 2 · answered Mar 11 '22 at 01:37

We want to show: $ \sqrt[n]{n} < \frac{ n + \sqrt{2n} } { n} $.

Recognizing this in it's AM-GM form, we want to find $n$ terms whose product is $n$, and whose sum is (at most) $ n + \sqrt{2n}$. With those, we can conclude that $$ \frac{ n + \sqrt{2n} } { n } \geq \frac{\sum x_i } { n} \geq \sqrt[n]{\prod x_i } = \sqrt[n]{n}.$$

There are various ways of finding such a set, especially when we restrict to large enough $n$.

EG

For $n \geq 3$, take $ x_1 = x_2 = x_3 = \sqrt[3]{n}, x_4 = \ldots = x_n = 1$.
Verify that $ n + \sqrt{2n} > 3 \sqrt[3]{n} + (n-3)$.

For $n =1, 2$, we can manually check that inequality holds.

Prove that $\,\sqrt [n] n < 1 + \sqrt{\frac{2}{n}}\,$

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