# Let $\left<a_i\right>,\left<b_i\right>$ be real sequences satisfied as following conditions：

$$\left, \left$$ 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 knew of one solution that was some crazy Sum of Squares (which I don't know how to reproduce). Mar 24, 2022 at 13:46
• Found a solution using Approach0 -> artofproblemsolving.com/community/c6h2012699p14113052 Mar 24, 2022 at 13:54
• wow, thank you very much！
– 余志祥
Mar 24, 2022 at 16:33
• But i was wondered that how to find solution. I google it so many days.
– 余志祥
Mar 24, 2022 at 16:36
• It was the first result in this Approach0 search. Mar 24, 2022 at 16:37

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 $$b$$".

• 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. Mar 24, 2022 at 18:54
• @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.) Mar 24, 2022 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 $$(e\cdot a)^2 + (e\cdot b)^2 \leq (\tilde{e}\cdot\tilde{a})^2 + (\tilde{e} \cdot \tilde{b})^2 = 1 \, .$$

• Ah, we came to the same approach :) Mar 24, 2022 at 18:18
• Found some more :P Mar 25, 2022 at 23:02
• Thank you all very much.
– 余志祥
Mar 27, 2022 at 2:13