[Toán 11] Dãy số

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truongduong9083

Tam giác ABC vuông tại A suy ra $a^2 = b^2+c^2 \Rightarrow (\dfrac{b}{a})^2+(\dfrac{c}{a})^2 = 1$
$\Rightarrow 0 < \dfrac{b}{a}<1; 0 < \dfrac{c}{a}<1 $
Với $n \geq 2 \Rightarrow (\dfrac{b}{a})^n \leq (\dfrac{b}{a})^2; (\dfrac{c}{a})^n \leq (\dfrac{c}{a})^2$
$\Rightarrow (\dfrac{b}{a})^n+(\dfrac{c}{a})^n \leq 1$
$\Leftrightarrow b^n+c^n \leq a^n$
 
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