Bất Đẳng Thức

Q

quangltm

$\sum \frac a{\sqrt{b^3+1}} = \sum \frac a{\sqrt{(b+1)(b^2-b+1)}} \overset{A-G}{\ge} \sum \frac {2a}{b^2+2}$
Ta cần chứng minh $2\sum \frac a{b^2+2} \ge 2$
Thật vậy:
$$2\sum \frac a{b^2+2} = \sum a - \sum \frac{ab^{2}}{\frac{b^2}{2}+\frac{b^2}{2}+2}\ge 6-\sum \frac{ab^{2}}{3b\sqrt[3]{\frac{b}{2}}}= 6-\sum\frac{ab}{3\sqrt[3]{\frac{b}{2}}}$$
Mà
$$\left ( \frac{ab}{\sqrt[3]{\frac{b}{2}}}+\frac{bc}{\sqrt[3]{\frac{c}{2}}}+\frac{ca}{\sqrt[3]{\frac{a}{2}}} \right )^3 \overset{Hölder}{\le} \left ( \sum \frac{ab}{\frac{b}{2}} \right )(ab+bc+ca)^{2}= 2(\sum a)(\sum ab)^{2}\le 2.6.12^{2}= 12^{3}$$
$\implies Q.E.D$
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom