“Với 4 số a, b, x, y ta có: $\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)\ge {{\left( ax+by \right)}^{2}}$. Dấu “=” xảy ra khi nào?”
$\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)\ge {{\left( ax+by \right)}^{2}}$
$\Leftrightarrow \left( {{a}^{2}}+{{b}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)-{{\left( ax+by \right)}^{2}}\ge 0$
$\Leftrightarrow {{\left( ax \right)}^{2}}+{{\left( ay \right)}^{2}}+{{\left( bx \right)}^{2}}+{{\left( by \right)}^{2}}-{{\left( ax \right)}^{2}}-2abxy-{{\left( by \right)}^{2}}\ge 0$
$\Leftrightarrow {{\left( ay \right)}^{2}}+{{\left( bx \right)}^{2}}-2abxy\ge 0$
$\Leftrightarrow {{\left( ay-bx \right)}^{2}}\ge 0$(luôn đúng)
Dấu “=” xảy ra khi ay = bx $\Leftrightarrow \frac{a}{x}=\frac{b}{y}.$