MHB Can you prove this inequality challenge?

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The inequality challenge involves proving that for any positive integer n greater than 1, the expression $\sqrt[n]{1+\dfrac{\sqrt[n]{n}}{n}}+\sqrt[n]{1-\dfrac{\sqrt[n]{n}}{n}}$ is less than 2. Participants discuss various approaches to tackle the proof, including using properties of limits and the behavior of the function as n increases. The discussion highlights the significance of analyzing the terms within the roots and their convergence. Several users share insights on applying mathematical inequalities and transformations to simplify the problem. Ultimately, the goal is to establish the validity of the inequality through rigorous mathematical reasoning.
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Prove that $\sqrt[n]{1+\dfrac{\sqrt[n]{n}}{n}}+\sqrt[n]{1-\dfrac{\sqrt[n]{n}}{n}}<2$ for any positive integer $n>1$.
 
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First note, that the function $f(x) = x^{\frac{1}{n}}$ is concave (downward) for $n = 2,3,4 …$.

Applying Jensens inequality for a concave function:

\[\frac{1}{N}\sum_{i=1}^{N}f(x_i)\leq f\left ( \frac{\sum_{i=1}^{N}x_i}{N} \right )\]

Equality holds if and only if $x_1 = x_2 = … = x_N$ or $f$ is linear.

In our case: $N = 2$ and $1+\frac{\sqrt[n]{n}}{n} \neq 1-\frac{\sqrt[n]{n}}{n}$ for all positive integers $n$.

Note also, that the case $n = 1$ implies, that $f$ is linear, which is why the case is omitted. Hence, we obtain:

\[\sqrt[n]{1+\frac{\sqrt[n]{n}}{n}} + \sqrt[n]{1-\frac{\sqrt[n]{n}}{n}} < 2\sqrt[n]{\frac{1+\frac{\sqrt[n]{n}}{n}+1-\frac{\sqrt[n]{n}}{n}}{2}} = 2\] q.e.d.