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

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.

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 $(0,2)$.
