Cho a + b + c = 4034 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{2}\)
Tính \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
K
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PT
3 tháng 1 2018
\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
\(=\left(\dfrac{a}{c+b}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{a+c+b}{c+b}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3\)
\(=4034.\dfrac{1}{2}-3=2014\)
3 tháng 1 2018
Guể?
\(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=\dfrac{4034}{2}=2017\)
\(\Rightarrow1+\dfrac{a}{c+b}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}=2017\)
\(\Rightarrow\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=2014\)
AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Lời giải:
\((a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)
$\Leftrightarrow 2018.\frac{1}{2018}=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$\Leftrightarrow 1=1+1+1+S$
$S=1-1-1-1=-2$
H
1 tháng 4 2021
Với cả 3 phần thì dấu "=" xảy ra tại a=b=c=1.
a) \(\dfrac{a}{1+b^2}=\dfrac{a\left(1+b^2\right)}{1+b^2}-\dfrac{ab^2}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\)
(Cosi) \(\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)
Tương tự : \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2};\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\)
\(\Rightarrow P\ge\left(a+b+c\right)-\dfrac{ab+bc+ca}{2}\ge\left(CS\right)\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{6}=3-\dfrac{3^2}{6}=\dfrac{3}{2}\)
b) \(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge\left(CS\right)1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)
Tương tự : \(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)
\(\Rightarrow P\ge3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\)
c)\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\left(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\right)+\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)
6 tháng 2 2022
Xét \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=126.16=2016\)
\(\Leftrightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=2016\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=2013\)
Vậy A = 2013
