Cho a,b,c 0, a+b+c=1. CMR:
$ \sqrt[n]{ \dfrac{2}{3}+a }+ \sqrt[n]{ \dfrac{2}{3} +b} + \sqrt[n]{ \dfrac{2}{3}+c }$ 3
1 BĐT thú vị
Started By math_galois, 07-01-2008 - 21:13
#1
Posted 07-01-2008 - 21:13
#2
Posted 07-01-2008 - 21:53
$VT \leq 3\sqrt[n]{\dfrac{\dfrac{2}{3}+a+\dfrac{2}{3}+b+\dfrac{2}{3}+c}{3}}=3$ ->đpcm
HTA
dont put off until tomorrow what you can do today
#3
Posted 08-01-2008 - 19:29
bạn có thể nói rõ tại sao lại như vậy ko ?
#4
Posted 08-01-2008 - 21:07
Áp dụng bdt này nhé
$\dfrac{a_1^k+a_2^k+....+a_n^k}{n}\geq (\dfrac{a_1+a_2+...+a_n}{n})^k$
$\dfrac{a_1^k+a_2^k+....+a_n^k}{n}\geq (\dfrac{a_1+a_2+...+a_n}{n})^k$
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