cho a+b+c=2016 và 1/a+b + 1/b+c + 1/c+a = 1/504
Tính S= a/b+c + b/c+a + c/a+b
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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 \(P=-2\)
ST
1
T
8 tháng 7 2019
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow\left(x;y;z\right)\) suy ra x, y, z >0 và x + y + z = 2016
BĐT \(\Leftrightarrow\frac{\frac{1}{yz}}{\frac{1}{x^2}\left(\frac{3}{y}+\frac{1}{z}\right)}+\frac{\frac{1}{zx}}{\frac{1}{y^2}\left(\frac{3}{z}+\frac{1}{x}\right)}+\frac{\frac{1}{xy}}{\frac{1}{z^2}\left(\frac{3}{x}+\frac{1}{y}\right)}\ge504\)
\(\Leftrightarrow\frac{x^2}{3z+y}+\frac{y^2}{3x+z}+\frac{z^2}{3y+x}\ge504\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel suy ra:
\(VT\ge\frac{\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\frac{x+y+z}{4}=\frac{2016}{4}=504\) (đpcm)
Đẳng thức xảy ra khi x = y = z = 672 hay \(a=b=c=\frac{1}{672}\)
ND
0
S
1
HP
27 tháng 4 2016
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow S=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{c+a}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)
\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2016.\frac{1}{90}-3=\frac{97}{5}\)
Vậy....................
30 tháng 12 2016
Vì \(a+b+c=2016\Rightarrow a=2016-\left(b+c\right);b=2016-\left(a+c\right);c=2016-\left(a+b\right)\)
Ta có:\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S=\frac{2016-\left(b+c\right)}{b+c}+\frac{2016-\left(a+c\right)}{a+c}+\frac{2016-\left(a+b\right)}{a+b}\)
\(S=\frac{2016}{b+c}-1+\frac{2016}{a+c}-1+\frac{2016}{a+b}-1\)
\(S=2016.\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(S=2016.\frac{1}{2016}-3\)
\(S=-2\)
PT
0
Lời giải:
\(\left\{\begin{matrix} a+b+c=2016\\ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{504}\end{matrix}\right.\)
\(\Rightarrow (a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2016.\frac{1}{504}=4\)
\(\Leftrightarrow \frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}=4\)
\(\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}=4\)
\(\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+1+1+1=4\)
\(\Leftrightarrow S+3=4\Leftrightarrow S=1\)
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