Setup/ rephrasing the question:
Consider the vectors $ a = ( a_1, \ldots, a_n), b = (b_1, \ldots , b_n)$. The conditions state that these are orthogonal unit vectors, $ \angle (a, b) = 90^\circ$.
Consider unit vector $c = ( \frac{1}{\sqrt{n} } , \ldots \frac{1}{\sqrt{n}} )$ and $-c$.
Since $ \angle (a, c) + \angle (a, -c) = 180^\circ$ either $ \angle (a, c) \leq 90^\circ$ or $ \angle (a, -c) \leq 90^\circ$. WLOG, let it hold for $ c$.
The inequality is equivalent to showing that:
$$ (a\cdot c) ^2 + ( b \cdot c)^2 \leq 1 \Leftrightarrow \cos^2 \angle (a, c) + \cos^2 \angle (b, c) \leq 1.$$
Proof: Let $ \alpha = \angle (a, c) $ and $ \beta = \angle (b, c)$.
We have $ 90^\circ = \angle (a, b) \leq \angle (a, c) + \angle (c, b) = \alpha + \beta $ and $ \angle(b, c) \leq \angle (b, a) + \angle (a, c) $, thus$ 90^\circ - \alpha \leq \beta \leq \ \alpha + 90^\circ$.
Recall from above that the choice of $c$ resulted in $ 0^\circ \leq \angle (a, c) \leq 90^\circ$, hence $ \cos^2 \beta \leq \sin^2 \alpha$.
Thus, $ \cos^2 \alpha + \cos^2 \beta \leq \cos^2 \alpha + \sin^2 \alpha = 1$ as desired.
Thus the inequality is true.
Equality holds iff the various "angle at a point inequality" holds, meaning that $a, b, c$ lie on the same plane (and $a, b$ are orthogonal as per the condition). Note that we do not require "$c$ lies in-between $a$ and $b$".
Best Answer
Apply the AM-GM inequality to the sequence $a_k/(a_k+b_k)$ and then to the sequence $b_k/(a_k+b_k)$. Add the resulting two inequalities, and multiply through by $\left(\prod_k (a_k+b_k)\right)^{1/n}$ to get the result.
This is exercise 2.11 (page 34) of J. Michael Steele's The Cauchy-Schwarz Master Class, and the result is there credited to Minkowski. This inequality is sometimes called the "superadditivity of the geometric mean".