Question

[tex] Fie~A=$\begin{pmatrix}
a & b & c \\
c & a & b \\
b & c& a
\end{pmatrix} $, a,b,c\in R, a^2+b^2+c^2=1.~Aratati~ca~|det(A)|\leq
1 [/tex]

Answer 1

$\displaystyle Asa~cum~iti~spuneam~in~privat,~problema~este~de-a~10-a,~dar \\ \\ deghizata.~Avem~\det A =a^3+b^3+c^3-3abc,~deci~problema~poate \\ \\ fi~reformulata~astfel: \\ \\ Fie~a,b,c \in \mathbb{R}~astfel~incat~a^2+b^2+c^2=1.~Demonstrati~ca \\ \\ |a^3+b^3+c^3-3abc| \le 2.$

$\displaystyle Folosim~identitatea~foarte~cunoscuta \\ \\ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac). \\ \\ Deci~a^3+b^3+c^3-3abc=(a+b+c)(1-ab-bc-ac). \\ \\ Notam~s=a+b+c~si~p=ab+bc+ac. \\ \\ Trebuie~sa~demonstram~ca~|s(1-p)| \le 1. \\ \\ Avem~s^2=1+2p. \\ \\ |s(1-p)|= \left | s \cdot \frac{3-s^2}{2} \right |. \\ \\ Deci~avem~de~demonstrat~ca~|3s-s^3| \le 2,~adica~ -2 \le 3s-s^3 \le 2.$

$\displaystyle Acest~lucru~se~poate~demonstra~usor~cu~ajutorul~analizei.~Voi \\ \\ lucra~fara~analiza,~totusi. \\ \\ Stim~ca~|s| \le \sqrt{3(a^2+b^2+c^2)}=\sqrt{3}.~Deci~s \in [- \sqrt 3, \sqrt 3]. \\ \\ \bullet Prima~parte: -2 \le 3s-s^3 \Leftrightarrow s^3-3s-2 \le 0 \Leftrightarrow \\ \\ \Leftrightarrow (s+1)^2(s-2) \le 0,~adevarat. \\ \\ \bullet A~doua~parte:~3s-s^3 \le 2 \Leftrightarrow s^3-3s+2 \ge 0 \Leftrightarrow \\ \\ (s-1)^2(s+2) \ge 0,~adevarat.$