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?

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