Biết (a+b-c)/c = (b+c-a)/a = (a+c-b)/b và a,b,c thuộc R*
Tính (1+b/a)(1+a/c)(a+c/b)
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VT
3 tháng 3 2019
Áp dụng BĐT svac, ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=\frac{9}{1}=9\left(ĐPCm\right)\)
Dấu = xảy ra <=> a=b=c=1/3
13 tháng 7 2019
Ta có: \(\frac{1}{x\left(a-b\right)\left(a-c\right)}+\frac{1}{y\left(b-a\right)\left(b-c\right)}+\frac{1}{z\left(c-a\right)\left(c-b\right)}\)
\(=\frac{1}{x\left(a-b\right)\left(a-c\right)}-\frac{1}{y\left(a-b\right)\left(b-c\right)}+\frac{1}{z\left(a-c\right)\left(b-c\right)}\)
\(=\frac{yz\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{xz\left(a-c\right)}{yxz\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{xy\left(a-b\right)}{zxy\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{yz\left(b-c\right)-xz\left(a-c\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)\(=\frac{yz\left(b-c\right)-xz\left[\left(b-c\right)+\left(a-b\right)\right]+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{yz\left(b-c\right)-xz\left(b-c\right)-xz\left(a-b\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(y-x\right)-\left(a-b\right)x\left(z-y\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(c+a-b-b-c+a\right)-\left(a-b\right)x\left(a+b-c-c-a+b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(2a-2b\right)-\left(a-b\right)x\left(2b-2c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)2z\left(a-b\right)-\left(a-b\right)2x\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(2z-2x\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{2\left(z-x\right)}{xyz\left(a-c\right)}=\frac{2\left(a+b-c-b-c+a\right)}{xyz\left(a-c\right)}\)
\(=\frac{2\left(2a-2c\right)}{xyz\left(a-c\right)}=\frac{2.2\left(a-c\right)}{xyz\left(a-c\right)}=\frac{4}{xyz}\Rightarrowđpcm\)
18 tháng 12 2019
Vì a + c = 2016 -> a = 2016 - [ b + c] ; b = 2016 - [ a + c] ; c = 2016 - [ a - b]
Ta có: S = a/ b + c + b/ a + c + c/a + b
S = 2016 - [ b + c] + 2016 - [ a + c] + 2016 - [ a + b]
S = 2016/ b + c - 1 + 2016/a + c - 1 + 2016/a + b
S = 2016.[ 1/b + c + 1/a + c + 1/a + b] - 3
S = 2016. 1/2016 - 3
S = - 2
CC
18 tháng 12 2019
Từ \(a+b+c=2016\) và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2016}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=2016.\frac{1}{2016}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=1\)
\(\Rightarrow\frac{\left(a+b\right)+c}{a+b}+\frac{\left(a+c\right)+b}{a+c}+\frac{\left(b+c\right)+a}{b+c}=1\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}=1\)
\(\Rightarrow\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=-2\)
hay \(P=-2\)
HA
0
VT
30 tháng 6 2018
Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)
VT
30 tháng 6 2018
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)
=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)
2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)
sửa đề:1+c/b chứ ko phải là a+c/b nhé bn
+)Xét a+b+c=0
=>a+b=-c;b+c=-a;c+a=-b
Khi đó \(\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{c+a}{c}\right)\left(\frac{b+c}{b}\right)\)
\(=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=-1\)
+)Xét a+b+c \(\ne\) 0
Theo t/c dãy....:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c+b+c-a+a+c-b}{c+a+b}=\frac{a+b+c}{c+a+b}=1\)
=>a+b-c=c=>a+b=2c
b+c-a=a=>b+c=2a
a+c-b=b=>a+c=2b
\(\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=\frac{2a.2b.2c}{a.b.c}=2.2.2=8\)
Vậy........................