Cho $k_1+k_2+...k_5=1$.Tìm Max của $A=k_1k_2+k_2k_3+k_3k_4+k_4k_5$
Cho $k_1+k_2+...k_5=1$.Tìm Max của $A=k_1k_2+k_2k_3+k_3k_4+k_4k_5$
Started By Minhnguyenthe333, 18-08-2015 - 22:08
#1
Posted 18-08-2015 - 22:08
- buibichlien likes this
#2
Posted 18-08-2015 - 22:19
$A=k_1.k_2+k_2.k_3+k_3.k_4+k_4.k_5 \leq (k_1+k_3+k_5)(k_2+k_4)$
$\sqrt{(k_1+k_3+k_5)(k_2+k_4)}\leq \frac{k_1+k_2+k_3+k_4+k_5}{2}=\frac{1}{2}$
$\rightarrow A\leq \frac{1}{4}$
- Minhnguyenthe333 and buibichlien like this
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users