$\left<a_i\right>, \left<b_i\right>$ be real sequences with
$$a_1^2+a_2^2+\cdots+a_n^2=1，\\
b_1^2+b_2^2+\cdots+b_n^2=1，\\
a_1b_1+a_2b_2+\cdots+a_nb_n=0.$$
Prove that $(a_1+a_2+\cdots+a_n)^2+(b_1+b_2+\cdots+b_n)^2\leq n$.
My attempt：
I try to prove it by induction，
As $n=2$，$a_1^2+a_2^2=1，b_1^2+b_2^2=1，a_1b_1+a_2b_2=0$
Because $a_1, a_2, b_1, b_2$ cannot be all $0$，W.L.O.G.，Assume $a_1\neq0$，
$b_1=-\frac{a_2b_2}{a_1}\Rightarrow$ $\frac{a_2^2b_2^2}{a_1^2}+b_2^2=1\Rightarrow b_2^2=a_1^2，b_1^2=a_2^2$
\begin{align}
(a_1+a_2)^2+(b_1+b_2)^2 & = a_1^2+a_2^2+b_1^2+b_2^2+2(a_1a_2+b_1b_2)\\
&=2+2(a_1a_2-\frac{a_2b_2^2}{a_1})\\
&=2+2\cdot\frac{a_2}{a_1}(a_1^2-b_2^2)\\
&=2
\end{align}
And the induction hypothesis is that $n=k$，$a_1^2+a_2^2+\cdots+a_k^2=1，b_1^2+b_2^2+\cdots+b_k^2=1，a_1b_1+a_2b_2+\cdots+a_kb_k=0，$
$(a_1+a_2+\cdots+a_k)^2+(b_1+b_2+\cdots+b_k)^2\leq k$ holds；
As $n=k+1$，$a_1^2+a_2^2+\cdots+a_k^2+a_{k+1}^2=1，b_1^2+b_2^2+\cdots+b_k^2+b_{k+1}^2=1，a_1b_1+a_2b_2+\cdots+a_kb_k+a_{k+1}b_{k+1}=0，$
Assume $a_1\neq 0$，
\begin{align}
& b_1=-\frac{a_2b_2+a_3b_3+\cdots a_{k+1}b_{k+1}}{a_1}\\
\Rightarrow & \frac{(a_2b_2+a_3b_3+\cdots+a_{k+1}b_{k+1})^2}{a_1^2}+b_2^2+\cdots+b_{k+1}^2=1\\
\Rightarrow &
(a_2b_2+a_3b_3+\cdots+a_{k+1}b_{k+1})^2+a_1^2b_2^2+\cdots+a_1^2b_{k+1}^2=a_1^2\\
\Rightarrow & …
\end{align}
I cannot finish the proof. Please help~
Best Answer
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 $b$".