$\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}$
Edited by Tham Lang, 01-09-2012 - 22:39.
Edited by Tham Lang, 01-09-2012 - 22:39.
Freedom Is a State of Mind
$\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}$
- tkvn 97-
Freedom Is a State of Mind
Ta áp dụng bđt Schawz$\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}$
Edited by duongchelsea, 01-09-2012 - 21:20.
$\sum \frac{1}{2a+b+c}\leq \frac{1}{4}(\sum \frac{1}{a+b}+\sum \frac{1}{a+c})\leq \frac{1}{8}(\sum \frac{1}{a})(Q.E.D)$$\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}$
TRIETHUYNHMATH
___________________________
08/12/1997
0 members, 1 guests, 0 anonymous users