Let $\left,\left$ be real sequences satisfied as following conditions： $\left<a_i\right>, \left<b_i\right>$ 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~
 A: 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 $b$".
A: Set $\vec{e}=(1/\sqrt{n},...,1/\sqrt{n})^t$, $\vec{a}=(a_1,...,a_n)^t$ and $\vec{b}=(b_1,...,b_n)^t$.
WTS: $$(\vec{e}\cdot \vec{a})^2 + (\vec{e}\cdot \vec{b})^2 \leq 1 \, . \tag{1}$$
Proof 1: Setting $\vec{e}\cdot \vec{b}=\cos\theta$, where $\theta$ is the angle between $\vec{b}$ and $\vec{e}$. Since $\vec{a}$ is orthogonal to $\vec{b}$, it lives in the hyperspace $B_\perp$ with normal vector $\vec{b}$. Since the normal vector of $B_\perp$ is tilted with respect to $\vec{e}$ by angle $\theta$, any projection of a unit-vector in $B_\perp$ onto $\vec{e}$ can be at most $|\sin\theta|$ in magnitude. This is the case when $\vec{a},\vec{b},\vec{e}$ lie in a common plane s.t.
$$\left|\angle(\vec{a},\vec{e}) \pm \angle(\vec{b},\vec{e})\right|=\angle(\vec{a},\vec{b})=90°\\
\text{or}\\
\angle(\vec{a},\vec{e}) + \angle(\vec{b},\vec{e}) = 270° \, .$$
Hence
$$(\vec{e}\cdot \vec{a})^2 + (\vec{e}\cdot \vec{b})^2 \leq \sin^2\theta + \cos^2\theta = 1 \, .$$

Proof 2: More formally, we write $$\vec{e}=R\vec{e}_1$$ for some rotation matrix $R$ and $\vec{e}_1=(1,0,...,0)^t$. Using this in (1), we find
$$\left(\vec{e}_1 \cdot R^t \vec{a}\right)^2 + \left(\vec{e}_1 \cdot R^t \vec{b}\right)^2 = \left(\vec{e}_1 \cdot \vec{a}'\right)^2 + \left(\vec{e}_1 \cdot \vec{b}' \right)^2 \, . \tag{2}$$
Since $R$ is orthogonal, $\vec{a}'$ and $\vec{b}'$ still satisfy all the conditions $$\vec{a}'^2=1 \\
\vec{b}'^2=1\\
\vec{a}'\cdot\vec{b}'=0 \, .$$
Using spherical coordinates, we can write $$\vec{b}'=\begin{pmatrix} \cos\theta \\ \sin\theta \cos\phi_1 \\ \sin\theta \sin\phi_1\cos\phi_2 \\
\vdots \\ \sin\theta \sin\phi_1 \cdots \sin\phi_{n-3} \cos\phi_{n-2} \\
\sin\theta \sin\phi_1 \cdots \sin\phi_{n-3} \sin\phi_{n-2} \end{pmatrix} \, .$$
Since $\vec{a}'$ is orthogonal to $\vec{b}'$, it has the form
$$\vec{a}'=c_\theta \, \partial_\theta \vec{b}' + \sum_{k=1}^{n-2} c_k \, \partial_{\phi_k} \vec{b}' = \begin{pmatrix} -c_\theta\sin\theta \\ c_\theta \cos\theta \cos\phi_1 - c_1 \sin\theta \sin\phi_1 \\ \vdots \end{pmatrix} \, .$$
Furthermore, since $\vec{a}'^2=1$ and the system of basis vectors is orthogonal, we have
$$1=\vec{a}'^2=c_\theta^2 + (\text{stuff } \geq 0) \\
\Rightarrow \quad c_\theta^2 \leq 1$$
Hence it is clear that following (2)
$$a_1'^2 + b_1'^2 = c_\theta^2 \sin^2\theta + \cos^2\theta \leq \sin^2\theta+\cos^2\theta = 1 \, .$$


Proof 3: I found this neat proof, using rotation matrices, that reduces the general case to the case $n=2$. Dropping vectors, $a,b,e$ are as above and assume $n\geq 3$. By rotation with $R$, we could write $e=Re_1$ and we arrived at (2), with $a',b'$ satisfying the conditions as before. We now rotate by $R'$ in the n-1 dimensional subspace s.t. $R'e_1=R'^te_1=e_1$. We define $R'$ s.t.
$$R'a'=R'\begin{pmatrix} a_1' \\ a_2' \\ a_3' \\ \vdots \end{pmatrix}=a''=\begin{pmatrix} a_1' \\ a_2'' \\ 0 \\ \vdots \end{pmatrix}$$
and we can thus write (2) as $$(R'^t e_1\cdot a')^2 + (R'^t e_1 \cdot b')^2 = (e_1\cdot R'a')^2 + (e_1 \cdot R'b')^2 = (e_1\cdot a'')^2 + (e_1 \cdot b'')^2 \, . \tag{3}$$
By construction $$a''^2=a_1'^2 + a_2''^2 = b''^2=b_1'^2 + b_2''^2 + b_3''^2 + ...=1 \quad , \quad a''\cdot b'' = a_1'b_1' + a_2''b_2'' = 0 \, ,$$
from which it is clear that $b_1'^2 + b_2''^2 \leq 1$. Now we define the rotation $R''$ by $$e_{12}=\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \\ \vdots\end{pmatrix} = R''e_1$$
i.e. $R''$ is a rotation by $45°$ in the 2-dimensional subspace spanned by $e_1,e_2$.
Thus, (3) then becomes
$$(R''^t e_{12} \cdot a'')^2 + (R''^t e_{12} \cdot b'')^2 = (e_{12} \cdot R''a'')^2 + (e_{12} \cdot R''b'')^2 = (e_{12} \cdot a''')^2 + (e_{12} \cdot b''')^2 \, . \tag{4}$$
Furthermore, we can now define
$$\tilde{a}=\begin{pmatrix} a_1''' \\ a_2''' \end{pmatrix} \quad , \quad \tilde{b}=\frac{1}{\sqrt{b_1'^2 + b_2''^2}}\begin{pmatrix} b_1''' \\ b_2''' \end{pmatrix} \quad , \quad \tilde{e}=\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \, ,$$
and because $R''$ is a rotation in the 2-dimensional subspace spanned by $e_1$ and $e_2$, it is plain that
$$\tilde{a}^2=a'''^2 = a''^2 = 1 = \tilde{b}^2 \quad , \quad \tilde{a}\cdot \tilde{b} = a''' \cdot b''' = a'' \cdot b'' = 0$$
and so, continuing with (4), it follows
$$(e\cdot a)^2 + (e\cdot b)^2 \leq (\tilde{e}\cdot\tilde{a})^2 + (\tilde{e} \cdot \tilde{b})^2 = 1 \, .$$
