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
Tru09
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Thiếu úy
- Oral1020 and nguyen tien dung 98 like this
#2
Posted 12-03-2013 - 17:04
Oral1020
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Thịnh To Tướng
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
vutuanhien
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Thiếu úy
$VT=a(1-b)+b(1-c)+c(1-d)+d(1-a)\leq \frac{a^2+1+b^2-2b}{2}+\frac{b^2+1+c^2-2c}{2}+\frac{c^2+1-2d+d^2}{2}+\frac{d^2+a^2-2a+1}{2}=(a^2+b^2+c^2+d^2)+2-(a+b+c+d)\leq 2$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
"The first analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months—when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado!" - Grothendieck
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