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V

vipboycodon

$\sqrt{a}+\sqrt{b} \le 2\sqrt{\dfrac{a+b}{2}}$
<=> $a+b+2\sqrt{ab} \le 4.\dfrac{a+b}{2}$
<=> $a+b+2\sqrt{ab} \le 2(a+b)$
<=> $2\sqrt{ab} \le a+b$
<=> $(\sqrt{a}-\sqrt{b})^2 \ge 0$ (đúng)
Dấu "=" xảy ra khi $a = b$.
 
C

congchuaanhsang

Theo Cauchy - Schwarz

$(\sqrt{a}+\sqrt{b})^2$\leq$2(a+b)$

\Leftrightarrow$\sqrt{a}+\sqrt{b}$\leq$\sqrt{2(a+b)}=2\sqrt{\dfrac{a+b}{2}}$
 
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