cho cac so duong a,b,c . Chung minh 1<a/a+b +b/b+c + c/c+a <2
Những câu hỏi liên quan
cho a,b,c la cac so duong thoa man (1/a+1/b+1/c)>=(a+b+c), chung minh a+b+c>=3abc
cho a,b,c la cac so duong thoa man (1/a+1/b+1/c)>=(a+b+c), chung minh a+b+c>=3abc
https://www.facebook.com/OnThiDaiHocKhoiA/posts/508217699295984
Đúng 0
Bình luận (0)
cho x,y,z la cac so nguyen duong va x+y+z la so le, cac so thuc a,b,c thoa man (a-b)/x=(b-c)/y=(a-c)/z. chung minh rang a=b=c
chung minh rang voi a,b,c la cac so duong ,ta co (a+b+c)(1/a+1/b+1/c)>=9
Ta có (a+b+c)(1/a+1/b+1/c) = 1 + 1 + 1 + a/b + a/c + b/a + b/c + c/a + c/b
= 3 + (a/b + b/a) + (a/c + c/a) + (b/c + c/b) (1)
Vì a, b, c > 0 nên ta có (Áp dụng Côsi)
a/b + b/a \(\ge\) 2 (2)
a/c + c/a \(\ge\) 2 (3)
b/c + c/b \(\ge\) 2 (4)
Từ (1), (2), (3) và (4) suy ra
(a+b+c)(1/a+1/b+1/c) \(\ge\) 9
Dấu "=" xảy ra <=> a = b = c
Đúng 0
Bình luận (0)
1. Cho a,b la 2 so duong thoa a+b<=1.chung minh rang \(6b+\frac{1}{3a}+\frac{4}{b}\ge11\).
2. cho a,b,c la cac so nguyen duong sao cho (a-b).(a-c).(b-c)=a+b+c
a. chung minh rang a+b+c chia het cho 2
b. Tim gia tri nho nhat cua M=a+b+c
voi a,b,c,d, la cac so duong thoa man a*b = c*d =1 chung minh bat dang thuc : ( a+b )*( c+d ) +4 >= 2*( a+b+c+d ) cac ban oi giup minh voi OK
Cho a,b,c la cac so duong a+b+c=3
Chung minh:\(a^5+b^5+c^5+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge6\)
Áp dụng bđt AM-GM:
\(a^5+\frac{1}{a}\ge2\sqrt{a^5.\frac{1}{a}}=2a^2\)
\(b^5+\frac{1}{b}\ge2\sqrt{b^5.\frac{1}{b}}=2b^2\)
\(c^5+\frac{1}{c}\ge2\sqrt{c^5.\frac{1}{c}}=2c^2\)
\(\Rightarrow VT\ge2\left(a^2+b^2+c^2\right)\ge\frac{2}{3}\left(a+b+c\right)^2=6\)
\("="\Leftrightarrow a=b=c=1\)
Đúng 0
Bình luận (0)
Voi a,b,c la cac so duong thoa man a*b =c*d =1 chung minh (a+b)(c+d) + 4>= 2(a+b+c+d)
Cho a,b,c,d la cac so duong sao cho a+b+c = 1 . Chung minh \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}\ge\frac{1}{2}\)