bất đẳng thức,hay

V

vy000

$\dfrac{bc}{\sqrt{a+bc}}=\dfrac{bc}{\sqrt{a^2+bc+ca+ab}}=\dfrac{bc}{\sqrt{(a+b)(a+c)}} \le \dfrac{bc}{2(a+b)}+\dfrac{bc}{2(a+c)}$

Chứng minh tương tự:

$\dfrac{ca}{\sqrt{b+ac}} \le \dfrac{ca}{2(b+c)}+\dfrac{ca}{2(b+a)}$
$\dfrac{ab}{\sqrt{c+ab}} \le \dfrac{ab}{2(a+c)}+\dfrac{ba}{2(b+c)}$

Cộng hết lại là xong :)
 
You must log in or register to reply here.
Top Bottom