Làm cụ thế được không
Ta co: $\frac{a}{b+c}< \frac{a+a}{a+b+c}< \frac{2a}{a+b+c}$
$\Leftrightarrow \sum \frac{a}{b+c}< \sum \frac{2a}{a+b+c}=2$
Vay $\sum \frac{a}{b+c}< 2$
Ta co: $\sqrt{a(b+c)}\leq \frac{a+b+c}{2}$
$\Leftrightarrow \frac{1}{\sqrt{a(b+c)}}\geq \frac{2}{a+b+c}$
$\Leftrightarrow \frac{\sqrt{a}}{\sqrt{b+c}}\geq \frac{2a}{a+b+c}$
$\Leftrightarrow \sum \frac{\sqrt{a}}{\sqrt{b+c}}\geq \sum \frac{2a}{a+b+c}=2$
BDT duoc chung minh
- anhtukhon1 yêu thích