chứng minh rằng

B

braga

Theo $AM-GM$ ta có:
$\dfrac{a^2}{b} + \dfrac{a^2}{b} + \dfrac{a^2}{b} + \dfrac{a^2}{b} + \dfrac{b^2}{2c} + \dfrac{b^2}{2c}+\dfrac{4c^2}{a}\ge 7a \\ \implies \dfrac{4a^2}{7b}+\dfrac{b^2}{7c}+\dfrac{4c^2}{7a} \ge a \ (1) \\ \text{Lại có:} \ \dfrac{a^2}{b} + \dfrac{b^2}{2c} + \dfrac{b^2}{2c} + \dfrac{b^2}{2c} + \dfrac{b^2}{2c} + \dfrac{4c^2}{a}+\dfrac{4c^2}{a}\ge 7b \\ \implies \dfrac{3a^2}{7b} + \dfrac{6b^2}{7c} +\dfrac{24c^2}{7a} \ge 3b \ (2)$
Cộng $(1)$ với $(2)\implies dpcm$
 
You must log in or register to reply here.
Top Bottom