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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ It was the first result in this Approach0 search. $\endgroup$
    – Calvin Lin
    Mar 24 at 16:37

2 Answers 2


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 $b$".

  • $\begingroup$ 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

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 $$(e\cdot a)^2 + (e\cdot b)^2 \leq (\tilde{e}\cdot\tilde{a})^2 + (\tilde{e} \cdot \tilde{b})^2 = 1 \, .$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.