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.
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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