Cho 4 số thực a,b,c,d không âm và nhỏ hơn hoặc bằng 1.
Chứng minh rằng $0 \leq a+b+c+d-ab-bc-cd-da \leq 2$
Khi nào đẳng thức xảy ra ?
CMR :$0 \leq a+b+c+d-ab-bc-cd-da \leq 2$
Started By Tru09, 12-03-2013 - 17:01
#1
Posted 12-03-2013 - 17:01
- Oral1020 and nguyen tien dung 98 like this
#2
Posted 12-03-2013 - 17:04
Ta có :
$a^2+b^2+c^2+d^2 \ge ab+bc+cd+da$
Do $a;b;c;d \in [0;1]$ nên
$a+b+c+d \ge a^2+b^2+c^2+d^2 \ge ab+bc+cd+da$
$\Longrightarrow a+b+c+d-ab-bc-cd-ad \ge 0$
$a^2+b^2+c^2+d^2 \ge ab+bc+cd+da$
Do $a;b;c;d \in [0;1]$ nên
$a+b+c+d \ge a^2+b^2+c^2+d^2 \ge ab+bc+cd+da$
$\Longrightarrow a+b+c+d-ab-bc-cd-ad \ge 0$
- IloveMaths and tramyvodoi like this
"If I feel unhappy,I do mathematics to become happy.
If I feel happy,I do mathematics to keep happy."
Alfréd Rényi
#3
Posted 12-03-2013 - 18:40
$VT=a(1-b)+b(1-c)+c(1-d)+d(1-a)\leq \frac{a^2+1+b^2-2b}{2}+\frac{b^2+1+c^2-2c}{2}+\frac{c^2+1-2d+d^2}{2}+\frac{d^2+a^2-2a+1}{2}=(a^2+b^2+c^2+d^2)+2-(a+b+c+d)\leq 2$Cho 4 số thực a,b,c,d không âm và nhỏ hơn hoặc bằng 1.
Chứng minh rằng $0 \leq a+b+c+d-ab-bc-cd-da \leq 2$
Khi nào đẳng thức xảy ra ?
Đẳng thức xảy ra khi a=c=1, b=d=0 hoặc b=d=1, a=c=0
- BlackSelena, Oral1020, IloveMaths and 1 other like this
"Algebra is the offer made by the devil to the mathematician. The devil says: I will give you this powerful machine, it will answer any question you like. All you need to do is give me your soul: give up geometry and you will have this marvelous machine." (M. Atiyah)
#4
Posted 22-08-2015 - 08:31
còn cách khác không?
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