Vì a/b<c/d=>a.d<c.b

<=>2018a.d<2018b.c

<=>2018a.d+cd<2018b.c+cd

<=>d(2018a+c)<c(2018b+d)

<=>điều phải chứng minh

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\) (do a,b,c >0)

Ta có đpcm

may hoc thay nghia a

#)Giải :

Ta có :

\(\hept{\begin{cases}\frac{a}{a+b}>\frac{a}{a+b+c}\\\frac{b}{b+c}>\frac{b}{a+b+c}\\\frac{c}{c+a}>\frac{c}{a+b+c}\end{cases}}\)

\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1\left(đpcm\right)\)

a/c +b/c+b/a+c/a+c/b+a/b>=6 ( ap dung cosi)

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}

Áp dụng bđt cô si cho 2 số dương ta có:

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)

=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc\cdot\frac{1}{abc}}=9\)

Cho a, b, c thuộc N bằng:

-, 5 < a < 10

a = 6, 7, 8, 9

-, 7 < c < 10

c = 8, 9

-, a < b < c

6 < 7 < 8 <9

Mình nghĩ vậy nhưng không biết đúng không, nếu đúng các bạn l i k e cho mình nhé

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(ab+ac+bc\right)\left(a+b+c\right)-9abc\ge0\)

\(\Leftrightarrow a^2b+a^2c+abc+abc+ab^2+b^2c+abc+ac^2+bc^2-9abc\ge0\)

\(\Leftrightarrow a^2b+a^2c+ab^2+b^2c+ac^2+bc^2-6abc\ge0\)

\(\Leftrightarrow\left(a^2b-2abc+bc^2\right)+\left(a^2c-2abc+b^2c\right)+\left(ab^2-2abc+ac^2\right)\ge0\)

\(\Leftrightarrow b\left(a-b\right)^2+c\left(a-c\right)^2+a\left(b-c\right)^2\ge0\)(luôn đúng \(\forall a;b;c>0\))

Vật bđt đã đc chứng minh

Cho a,b,c>0 thì dễ thôi :v

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}\)

Khi a=b=c