Cho a+b+c=\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)=3 Tính: \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
Cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
a+b+c = 2010 => a+b=2010-c ; b+c=2010-a ; c+a=2010-b
=> S = a/2010-a + b/2010-b + c/2010-c = 2010/2010-a - 1 + 2010/2010-b -1 + 2010/2010-c - 1
= 2010/b+c - 1 + 2010/c+a - 1 + 2010/a+b - 1
= 2010.(1/b+c + 1/c+a + 1/a+b) - 3
= 2010.1/3 - 3 = 667
Vậy S = 667
Tk mk nha
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2010\cdot\frac{1}{3}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2010}{3}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2010}{3}\)
\(\Rightarrow S+3=\frac{2010}{3}\)
\(\Rightarrow S=\frac{2010}{3}-3=\frac{2001}{3}=667\)
Ta có \(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
=\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{2010}{3}=670\)
\(\Rightarrow S=667\)
Cho a,b,c > 0.Chứng minh rằng
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)\(\frac{2}{a+b}\)+\(\frac{2}{b+c}\)+\(\frac{2}{c+a}\)
b,\(\frac{4}{a}\)+\(\frac{5}{b}\)+\(\frac{3}{c}\)\(\ge\)\(4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Cho a,b,c > 0.Chứng minh rằng
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)\(\frac{2}{a+b}\)+\(\frac{2}{b+c}\)+\(\frac{2}{c+a}\)
b,\(\frac{4}{a}\)+\(\frac{5}{b}\)+\(\frac{3}{c}\)\(\ge\)\(4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
Cho a+b+c=2028 và\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính giá trị nhỏ nhất của : \(Q=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Ta có:\(\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\left(a+b+c\right)=\frac{1}{3}.2028\)
=>\(\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{c+a}{c+a}+\frac{b}{c+a}\right)=676\)
=>\(\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}+3=676\)
=>\(Q=673\)
Vậy Q=673
dự đoán của chúa Pain
a=b=c=\(\frac{2028}{3}\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{2\left(a+b+c\right)}\left(cosi\right).\)
\(Q\ge\frac{\left(a+b+c\right)}{2\left(a+b+c\right)}+\frac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{2\left(a+b+c\right)}\)
\(Q\ge\frac{1}{2}+\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\left(a+b+c\right)}\)
có
\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge3\sqrt[3]{\sqrt{a^2b^2c^2}}=3\sqrt[3]{abc}\)
có
\(a+b+c\ge3\sqrt[3]{abc}\)
thay vào ta được
\(Q\ge\frac{1}{2}+\frac{3\sqrt[3]{abc}}{3\sqrt[3]{abc}}=\frac{1}{2}+1=\frac{3}{2}\)
dấu = xảy ra khi \(a=b=c=\frac{2028}{3}=676\)
thử thay vào ta được
\(Q=\frac{676}{2\left(676\right)}+\frac{676}{2\left(676\right)}+\frac{676}{2\left(676\right)}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\) ( đúng )
Cho a + b + c = 100 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính giá trị biểu thức : \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
tính \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{2010-\left(b+c\right)}{b+c}+\frac{2010-\left(c+a\right)}{c+a}+\frac{2010-\left(a+b\right)}{a+b}\)
\(=\frac{2010}{b+c}-\frac{b+c}{b+c}+\frac{2010}{a+b}-\frac{a+b}{a+b}+\frac{2010}{a+c}-\frac{a+c}{a+c}=\left(\frac{2010}{b+c}+\frac{2010}{a+b}+\frac{2010}{a+c}\right)-\left(1+1+1\right)\)
\(=2010.\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)-3=2010.\frac{1}{3}-3=670-3=667\)
Cho a,b,c là 3 số thực thõa mãn \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{4}{a+b+c}\)
Tính \(M=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)
cho ba số a,b,c thỏa mãn a+b+c =6 và \(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=\frac{3}{2}\).Tính giá trị của biểu thức \(P=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c=a}\)
Ta có :
\(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=\frac{3}{2}\)
\(\Leftrightarrow\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}+1=\frac{3}{2}+3\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)
\(\Leftrightarrow6.\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)
\(\Rightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{9}{2}:6=\frac{3}{4}\)
Vậy \(P=\frac{3}{4}\)
Bài 1 : Cho \(a+b+c=2007\)và\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{90}\)
Tính \(S=\frac{a}{b+C}+\frac{b}{c+a}+\frac{c}{a+b}\)
Bài 2 : Cho \(abc\ne0v\text{à}\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
Tính \(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Bài 3 : Cho \(a+b+c\ne0\)
Thoả mãn : \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)Tính \(P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)
cái này chắc k ai làm đâu. mệt lắm