Question

Cum se demonstreaza inegalitatea lui Cauchy-Buniakovski-Schwarz: (ax+by)²≤(a²+b²)(x²+y²)

Answer 1

$\displaystyle Calcul~direct. \\ \\ (ax+by)^2 \le (a^2+b^2)(x^2+y^2) \Leftrightarrow \\ \\ \Leftrightarrow a^2x^2+2abxy+b^2y^2 \le a^2x^2+a^2y^2+b^2x^2+b^2y^2 \Leftrightarrow \\ \\ \Leftrightarrow 2abxy \le a^2y^2+b^2x^2 \Leftrightarrow \\ \\ \Leftrightarrow 0 \le a^2y^2-2abxy+b^2y^2 \Leftrightarrow \\ \\ 0 \le (ay-bx)^2,~adevarat!$