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$\left<a_i\right>, \left<b_i\right>$ 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~

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  • $\begingroup$ I knew of one solution that was some crazy Sum of Squares (which I don't know how to reproduce). $\endgroup$
    – Calvin Lin
    Mar 24 at 13:46
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    $\begingroup$ Found a solution using Approach0 -> artofproblemsolving.com/community/c6h2012699p14113052 $\endgroup$
    – Calvin Lin
    Mar 24 at 13:54
  • $\begingroup$ wow, thank you very much! $\endgroup$
    – 余志祥
    Mar 24 at 16:33
  • $\begingroup$ But i was wondered that how to find solution. I google it so many days. $\endgroup$
    – 余志祥
    Mar 24 at 16:36
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    $\begingroup$ It was the first result in this Approach0 search. $\endgroup$
    – Calvin Lin
    Mar 24 at 16:37

2 Answers 2

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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 $b$".

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  • $\begingroup$ Just wondering, if $\beta$ is close to $90°$, then $\alpha$ can actually be close to $180°$, i.e. $\alpha+\beta\leq 270°$. It's the other extreme to $180°-\alpha$ where where $a$ and $c$ are amost antiparallel. $\endgroup$
    – Diger
    Mar 24 at 18:54
  • $\begingroup$ @Diger True, though that condition isn't needed and has since been removed. The key aspect is that $ 90 - \alpha \leq \beta \leq 90 + \alpha$, which is what allows us to bound $\cos \beta$. (This went through several iterations before I settled on this final version, just didn't clean up thoroughly enough.) $\endgroup$
    – Calvin Lin
    Mar 24 at 20:40
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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 $$(e\cdot a)^2 + (e\cdot b)^2 \leq (\tilde{e}\cdot\tilde{a})^2 + (\tilde{e} \cdot \tilde{b})^2 = 1 \, .$$

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  • $\begingroup$ Ah, we came to the same approach :) $\endgroup$
    – Calvin Lin
    Mar 24 at 18:18
  • $\begingroup$ Found some more :P $\endgroup$
    – Diger
    Mar 25 at 23:02
  • $\begingroup$ Thank you all very much. $\endgroup$
    – 余志祥
    Mar 27 at 2:13

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