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## SEEMOUS 2011 Problem 3

Given vectors $\overline{a},\overline{b},\overline{c} \in \Bbb{R}^n$, show that
$(\| \overline{a}\|\langle \overline{b},\overline{c}\rangle)^2+(\| \overline{b}\|\langle \overline{a},\overline{c}\rangle)^2\leq \|\overline{a}\|\|\overline{b}\|(\|\overline{a}\|\|\overline{b}\|+|\langle \overline{a},\overline{b}\rangle |)\|\overline{c}\|^2$, with the usual notations.
SEEMOUS 2010 Problem 3

Solution: First, reduce the problem to $\Bbb{R}^3$. How we can do that? Consider an orthonormal basis for which the first three vectors span the vector space spanned by $a,b,c$. An orthonormal transformation preserves the norm and the inner product, therefore, it would be enough to prove the inequality in the new basis, a basis in which $a,b,c$ have only the first three components non-zero. Therefore, if we solve the problem for $a,b,c \in \Bbb{R}^3$ we are done. But in $\Bbb{R}^3$ we can reduce the problem to a geometric one. First, note that changing the sign of some of the vectors $a,b,c$ does not change anything in the inequality. If we denote $\alpha=\angle (b,c),\ \beta=\angle (a,c),\ \gamma=\angle (a,b)$. Arrange the signs such that $\alpha ,\beta \in [0,\pi/2]$ and drawing a trihedron angle $OABC$ with supports of its sides determined by $a,b,c$ the projection of $C$ on the plane $OAB$ would be inside the angle $\angle OAB$.

It is easy to prove that if $x,y,z$ satisfy $x+y=z$ then $\cos^2 x + \cos^2 y+\cos^2 z-2\cos x\cos y\cos z=1$. The inequality we want to prove is $\cos^2 \alpha +\cos^2 \beta\leq 1 + |\cos \gamma|$. Note that if one of $\alpha,\beta$ is $\pi/2$ then we are done.

Project $C$ on the plane $OAB$ in point $P$, and denote $\alpha_1=\angle POB,\beta_1=\angle POA$. Project $P$ on $OA,OB$ in $A_1,B_1$. By expressing cosine in a right triangle, we get $\cos^2 \alpha=OB_1^2/OC^2\leq OB_1^2/OP^2=\cos^2 \alpha_1$. Similar to this we get $\cos^2 \beta \leq \cos^2 \beta_1$. Use the given identity for $\alpha_1+\beta_1=\gamma$ and get $\cos^2 \alpha_1+\cos^2\beta_1 =1+2 \cos \alpha_1 \cos \beta_1 \cos \gamma -\cos^2 \gamma$. Let’s prove that

$latex 1+2\cos\alpha_1\cos\beta_1\cos\gamma-\cos^2\gamma$ $\leq 1 +\cos \gamma$. Of course, if $\cos \gamma=0$ we are done. Else, the inequality is equivalent to $2 \cos \alpha_1 \cos \beta_1 \leq 1+\cos \gamma=\cos(\alpha_1+\beta_1)=1+\cos\alpha_1 \cos \beta_1 -\sin \alpha_1 \sin \beta_1$. Finally, this last inequality is equivalent to $\cos(\alpha_1 -\beta_1) \leq 1$, which is true.

This solves the problem for $n \geq 3$. For $n=1$ the inequality is trivial, and for $n=2$ we do something similar to the arguments above, but we are no longer in space, and in the plane, the problem is even easier.

1. March 9, 2011 at 11:11 am

Eu tot geometric ma gandisem. Si tot ca tine am presupus ca $\alpha,\beta\in [0,\pi /2]$. Trebuie sa aratam ca $|\cos\gamma |\geq\cos^2\alpha +\cos^2\beta -1=\frac 12(\cos 2\alpha +\cos 2\beta )=\cos (\alpha+\beta )\cos (\alpha -\beta )$. Avem $|\alpha -\beta |\leq\pi /2$ deci $\cos (\alpha -\beta )\geq 0$ deci putem presupune ca $\cos (\alpha+\beta ),\cos (\alpha -\beta )>0$. Insa $\alpha,\beta,\gamma$ sunt unghirile triedrului $0abc$ (care poate fi si degenerat) deci avem $0\leq\gamma\leq\alpha +\beta\leq\pi$. Rezulta ca $0<\cos (\alpha +\beta )\leq\cos\gamma$, care, impreuna cu $0<\cos (\alpha -\beta )\leq 1$ implica $\cos (\alpha+\beta )\cos (\alpha -\beta )\leq\cos\gamma$.

• March 9, 2011 at 12:36 pm

E adevarat ca la problema asta s-au dat mai multe solutii distincte? Asa am auzit.

2. March 12, 2011 at 12:31 am

O solutie foarte rapida ar fi sa scriem determinantul Gramm pentru 3 vectori oarecare a,b,c avand dimensiunea n. Stim ca are determinantul mai mare sau egal decat 0 si se obtine imediat concluzia.

(Enghish: A fast solution would be to write the Gramm determinant for the three vectors.)

3. March 16, 2011 at 10:58 am

Ce stiti despre studentul care a fost descalificat pentru ca a dat solutiile din lista scurta?

• March 16, 2011 at 11:49 am

Nu am auzit despre acest incident. Daca intr-adevar au fost scurgeri de informatii de la profesori catre studenti, acest fapt este foarte trist.

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