Cezar Lupu wrote: For $n\in\mathbb{N}$, $n\geq 2$, $a_{i}, b_{i}\in\mathbb{R}$, $1\leq i\leq n$, such that \[\sum_{i=1}^{n}a_{i}^{2}=\sum_{i=1}^{n}b_{i}^{2}=1, \sum_{i=1}^{n}a_{i}b_{i}=0. \] Prove that \[\left(\sum_{i=1}^{n}a_{i}\right)^{2}+\left(\sum_{i=1}^{n}b_{i}\right)^{2}\leq n. \] Let $\vec a(a_{1},a_{2},...a_{n}),$ $\vec b(b_{1},b_{2},...,b_{n}),$ $\vec c(\frac{1}{\sqrt n},\frac{1}{\sqrt n},...,\frac{1}{\sqrt n}),$ $(\widehat{\vec a,\vec c})=\alpha$ and $(\widehat{\vec b,\vec c})=\beta.$ Hence, $\vec a\cdot\vec b=0,$ $|\vec a|=|\vec b|=|\vec c|=1.$ Thus, it remains to prove that $\cos^{2}\alpha+\cos^{2}\beta\leq1,$ which triviality for $n=2$ and $n=3.$ Nice problem, Cezar.

Let $< \ldots , \ldots >$ the standard scalar product in $\mathbb{R}^{n}$. Let's call our vectors.. $\vec a = (a_{1}, ... ,a_{n})$ $\vec b= (b_{1}, ... ,b_{n})$ $\vec c=(1, 1, .... ,1)$ The hypotesis are $<\vec a , \vec a > = <\vec b, \vec b > = 1$ and $<\vec a, \vec b\geq 0$ Now we call $\vec r = \vec c-<\vec c , \vec a> \vec a-<\vec c , \vec b> \vec b$. Now we know $< \vec r , \vec r > \geq 0$. Showing calculation we see: \[<(\vec c-<\vec c , \vec a> \vec a-<\vec c , \vec b> \vec b),(\vec c-<\vec c , \vec a> \vec a-<\vec c , \vec b> \vec b)\geq \] \[= <\vec c, \vec c>+<\vec v , \vec a >^{2}<\vec a , \vec a>+<\vec v , \vec b >^{2}<\vec b , \vec b>+2<\vec v , \vec a > <\vec v , \vec b ><\vec a , \vec b>-2 <\vec v , \vec a > <\vec v , \vec a>-2 <\vec v , \vec b > <\vec v , \vec b> = \] \[= n-<\vec v , \vec a >^{2}-<\vec v , \vec b >^{2}\] i.e. $n \geq (a_{1}+....+a_{n})^{2}+(b_{1}+...+b_{n})^{2}$ i'm sorry $\vec v = \vec c$ :P

Nice solution,Simo_the_Wolf! Here is another but less beautiful solution: Let \[ A = \sum{a_i},p = \sqrt {n - A^2},B = \sum{b_i} \] WLOG,we assume $ B\ge 0$ what we need to prove is equivlent to $ B\le p$,we assume that $ p\not = 0$(otherwise $ p = 0$ ,then $ a_i = \frac {1}{\sqrt {n}}$,then $ B = 0 = p$) And we have the following identity: \[ B - \frac {p}{2} = B - \frac {p}{2}\sum{b_i} - A\sum{a_ib_i} = \sum{b_i - 2pb_i^2 - Aa_ib_i} = \sum{( - \frac {2}{p}(\frac {p}{2} b_i - \frac { - 1 + Aa_i}{2})^2 + \frac {(1 - Aa_i)^2}{2p}) = \sum{ - \frac {2}{p}(\frac {p}{2} b_i - \frac { 1 + Aa_i}{2} )^2 + \frac {p}{2} }}\] Hence \[ B\le p \]

Hawk Tiger wrote: Nice solution,Simo_the_Wolf! Here is another but less beautiful solution: Let \[ A = \sum{a_i},p = \sqrt {n - A^2},B = \sum{b_i} \] WLOG,we assume $ B\ge 0$ what we need to prove is equivlent to $ B\le p$,we assume that $ p\not = 0$(otherwise $ p = 0$ ,then $ a_i = \frac {1}{\sqrt {n}}$,then $ B = 0 = p$) And we have the following identity: \[ B - \frac {p}{2} = B - \frac {p}{2}\sum{b_i} - A\sum{a_ib_i} = \sum{b_i - 2pb_i^2 - Aa_ib_i} = \sum{( - \frac {2}{p}(\frac {p}{2} b_i - \frac { - 1 + Aa_i}{2})^2 + \frac {(1 - Aa_i)^2}{2p}) = \sum{ - \frac {2}{p}(\frac {p}{2} b_i - \frac { 1 + Aa_i}{2} )^2 + \frac {p}{2} }}\] Hence \[ B\le p \] Second equaution is wrong, isn't it?

Nice problem indeed ! Here's one of my proofs :

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And here it is the optimal proof ( elementary ) for this pearl :

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Leonardg Very nice solution

Nice....

IMO2022Goldinshallah wrote: Leonardg Very nice solution Moubinool wrote: Nice.... Thank you, my friends.

There exist two more methods (involving functional analysis)

@above, can you share them here?

@above, just use the operator norm

Here is the "continuous" version of this problem. Basically, both versions just say that the sum of the squared projections of two orthogonal vectors $a,b$ on any unit vector $u$ is at most $1$. This is true for any Hilbert space, not just $\mathbb{R}^n$ or $L^2$. The simplest idea (in my opinion) is to represent $a$ (resp. $b$) as a linear combination of $u$ and some other vector that is orthogonal to $u$, that is, to decompose $a,b$ with respect to some orthogonal basis, that includes $u$. As done in the link.

Very nice, professor