Cho a, b, c > 0
C/m: \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\)
Cho \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\) . Tính
\(M=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{a+d}{b+c}\)
cho \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)
tính M=\(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}=\frac{a+d}{b+c}\)
Cộng thêm 1 vào mỗi đẳng thức, ta được:
\(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
Vì các tử số của mỗi tỉ số bằng nhau nên các mẫu số của mỗi tỉ số cũng bằng nhau
\(\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow M=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{a+d}{b+c}=1+1+1+1=4\)
cho các số dương a, b, c. C/m \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
mau nha cần gấp
Cho \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
Tính \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}-\left(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\right)\)
trừ mỗi tỉ lệ cho 1 ta được:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{2a+b+c+d}{a}-\frac{a}{a}=\frac{a+2b+c+d}{b}-\frac{b}{b}=\frac{a+b+2c+d}{c}-\frac{c}{c}=\frac{a+b+c+2d}{d}-\frac{d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+Nếu a+b+c+d\(\ne\)0 thì a=b=c=d lúc đó
M=1+1+1+1=4
+Nếu a+b+c+d=0 thì a+b=-(c+d);b+c=-(d+a);c+d=-(a+b);d+a=-(b+c) lúc đó:
M=(-1)+(-1)+(-1)+(-1)=-4
\(\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{a+b+2c+d+a+b+c+2d}{c+d}=\frac{2a+2b+3c+3d}{c+d}\)
\(=\frac{2\left(a+b\right)}{c+d}+\frac{3\left(c+d\right)}{c+d}=2.\frac{a+b}{c+d}+3\)
\(\frac{2a+b+c+d}{a}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+b+c+2d}{a+d}=\frac{3a+3d+2c+2b}{a+d}\)
\(=\frac{3\left(a+d\right)}{a+d}+\frac{2\left(b+c\right)}{a+d}=3+2.\frac{b+c}{a+d}\)
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{2a+b+c+d+a+2b+c+d}{a+b}=\frac{3a+3b+2c+2d}{a+b}\)
\(=\frac{3\left(a+b\right)}{a+b}+\frac{2\left(c+d\right)}{a+b}=3+\frac{c+d}{a+b}.2\)
\(\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+2b+c+d+a+b+2c+d}{b+c}=\frac{3b+3c+2a+2d}{b+c}\)
\(=\frac{3\left(b+c\right)}{b+c}+\frac{2\left(a+d\right)}{b+c}=3+\frac{a+d}{b+c}.2\)
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
\(\Rightarrow\frac{2a+b+c+d}{a}+\frac{a+2b+c+d}{b}+\frac{a+b+2c+d}{c}+\frac{a+b+c+2d}{d}=5.4=20\)
\(\Rightarrow3+\frac{a+b}{c+d}.2+3+\frac{b+c}{a+d}.2+3+\frac{c+d}{a+b}.2+3+\frac{d+a}{b+c}.2=20\)
\(\Rightarrow2.\left(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\right)=20-3-3-3-3\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{b+a}+\frac{d+a}{b+c}=8:2=4\)
vậy \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=4\)
Cho a,b,c,d là các số dương . Tìm GTNN của biểu thức :
\(M=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d}\)
cái này mà là của lớp 3 à. Sao khó thế
cái này ít nhất cũng phải lớp 6 lớp 7
Đặt \(S=\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\)
\(=\frac{a^2}{ab+ac+ad}+\frac{b^2}{bc+bd+ab}+\frac{c^2}{cd+ac+bc}+\frac{d^2}{ad+bd+cd}\)
Theo Svac-xơ thì \(S\ge\frac{\left(a+b+c+d\right)^2}{2\left(ab+ac+ad+bc+bd+cd\right)}\)
\(=\frac{a^2+b^2+c^2+d^2+2\left(ab+ac+ad+bc+bd+cd\right)}{2\left(ab+ac+ad+bc+bd+cd\right)}\)
Ngoài ra ta có : \(\hept{\begin{cases}a^2+b^2\ge2ab;a^2+c^2\ge2ac;a^2+d^2\ge2ad\\b^2+c^2\ge2bc;b^2+d^2\ge2bd;c^2+d^2\ge2cd\end{cases}}\)
\(\Rightarrow3\left(a^2+b^2+c^2+d^2\right)\ge2\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Rightarrow S\ge\frac{\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)}{2\left(ab+ac+ad+bc+bd+cd\right)}=\frac{8}{6}=\frac{4}{3}\)
Đặt \(P=\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d}\)
\(=\frac{b}{a}+\frac{c}{a}+\frac{d}{a}+\frac{c}{b}+\frac{d}{b}+\frac{a}{b}+\frac{d}{c}+\frac{a}{c}+\frac{b}{c}+\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\)
\(=\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{d}{a}+\frac{a}{d}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{d}{b}+\frac{b}{d}\right)+\left(\frac{c}{d}+\frac{d}{c}\right)\)
\(\ge2.6=12\)
\(\Rightarrow M=S+P\ge\frac{5}{6}+12=12\frac{5}{6}\)
Dấu "=" xảy ra khi a = b = c = d
Cho: \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
Tính M= \(\frac{a+b}{c+d}=\frac{b+c}{d+a}=\frac{c+d}{a+b}=\frac{d+a}{b+c}\)
1.Cho
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)tính \(M=\frac{a+b}{c+d}+\frac{b+c}{a+d}=\frac{c+d}{a+b}=\frac{d+a}{b+c}\)
Ta có : \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)
<=> \(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
<=> \(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
Nếu a + b + c + d = 0
=> a + b = -(c + d)
b + c = -(a + d)
c + d = -(a + b)
d + a = -(b + c)
Khi đó M = \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Nếu a + b + c + d \(\ne\)0
=> \(\frac{1}{b+c+d}=\frac{1}{a+c+d}=\frac{1}{a+b+d}=\frac{1}{a+b+c}\)
=> b + c + d = a + c + d = a + b + d = a + b + c
=> a = b = c = d
Khi đó M = \(\frac{a+b}{c+d}+\frac{b+c}{a+b}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
Vậy nếu a + b + c + d \(\ne\)0 => M = 4
nếu a + b + c + d = 0 => M = -4
Cho a,b,c>0
Tìm GTNN : \(A=20\left(\frac{b+c}{a}+\frac{a+b}{c}+\frac{a+c}{b}\right)-17\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
Ta chứng minh bất đẳng thức phụ:
\(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge\frac{4a}{b+c}+\frac{4b}{c+a}+\frac{4c}{a+b}\). (*)
Thật vậy, áp dụng bất đẳng thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) với x, y > 0 ta có:
\(\frac{4a}{b+c}+\frac{4b}{c+a}+\frac{4c}{a+b}\le a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\).
Do đó (*) đúng.
Suy ra: \(A\ge80\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-17\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=63\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\).
Áp dụng bất đẳng thức \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\) (bất đẳng thức Nesbitt) ta có \(A\ge\frac{189}{2}\).
Đẳng thức xảy ra khi a = b = c.
Vậy Min A = \(\frac{189}{2}\) khi a = b = c.
Cho a,b,c > 0 . C/m \(\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}>=\frac{a+b+2c}{a+ab+c}+\frac{2a+3b+3c}{a+b+2c}\)