Prove that $\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2$

Question

Prove that $$\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2 \tag{1} $$ $\forall$ $n \gt 1$

I tried using Induction:

For the Base Step $n=2$ we have:

$$x=\sqrt{1+\frac{1}{\sqrt{2}}}+\sqrt{1-\frac{1}{\sqrt{2}}}$$ Then we get:

$$x^2=2+\sqrt{2}\lt 4$$

So $x \lt 2$

Now Let $P(n)$ is True, We shall need to prove $P(n+1)$ is also True

We have $P(n+1)$ as:

$$\left ( 1+\frac{(n+1)^{\frac{1}{n+1}}}{n+1} \right )^\frac{1}{n+1}+\left ( 1-\frac{(n+1)^{\frac{1}{n+1}}}{n+1} \right )^\frac{1}{n+1}$$

Now i tried to use the fact that:

$$f(x)=x^{\frac{1}{x}}$$ is a Monotone Decreasing $\forall x \ge e$

Hence $\forall n \ge 3$ we have:

$$(n+1)^{\frac{1}{n+1}} \lt n^{\frac{1}{n}} \tag{2}$$ and also

$$\frac{1}{n+1} \lt \frac{1}{n} \tag{3}$$

Multiplying $(2),(3)$ We get:

$$1+\frac{(n+1)^{\frac{1}{n+1}}}{n+1}\lt 1+\frac{n^{\frac{1}{n}}}{n}$$

Can we proceed from here?

Answer 1

Since $f(x)=x^{\frac{1}{n}}$ is a concave function for $n>1$, by Jensen we obtain: $$\left (1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}\leq2\left (\frac{1+\frac{n^{\frac{1}{n}}}{n}+ 1-\frac{n^{\frac{1}{n}}}{n}}{2}\right )^\frac{1}{n}=2.$$ The equality occurs for $$1+\frac{n^{\frac{1}{n}}}{n}=1-\frac{n^{\frac{1}{n}}}{n},$$ which says that the equality does not occur.

Answer 2

Note that by Bernoulli inequality in the form

$$(1+x)^a<1+ax, \quad 0<a<1$$

which can be easily proved by induction, we obtain

$$\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<1+\frac{n^{\frac{1}{n}}}{n^2}+1-\frac{n^{\frac{1}{n}}}{n^2} =2$$

Answer 3

By the generalized binomial formula, for $x\ne0$,

$$(1+x)^{1/n}+(1-x)^{1/n} \\=2+\frac2n\left(\frac1n-1\right)\frac{x^2}2+\frac2n\left(\frac1n-1\right)\left(\frac1n-2\right)\left(\frac1n-3\right)\frac{x^4}{4!}+\cdots\\<2.$$

That's all folks.

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Alternatively, the first derivative of $(1+x)^{1/n}+(1-x)^{1/n}$ is odd and monotonic in $(-1,1)$ because the second derivative is non-negative, so it has a single maximum, at $(0,2)$.