Can you prove this inequality challenge?

[anemone]

Prove that $\sqrt[n]{1+\dfrac{\sqrt[n]{n}}{n}}+\sqrt[n]{1-\dfrac{\sqrt[n]{n}}{n}}<2$ for any positive integer $n>1$.

 

[lfdahl]

Spoiler: Suggested solution

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.