\(\Leftrightarrow\dfrac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\le1\)

\(\Leftrightarrow\dfrac{ab+bc+ca+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\le1\)

\(\Leftrightarrow ab+bc+ca+12\le2\left(ab+bc+ca\right)+9\)

\(\Leftrightarrow ab+bc+ca\ge3\)

Hiển nhiên đúng do: \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}=3\)

Vì abc=1 , ta đặt \(a=\dfrac{x}{y};b=\dfrac{y}{z};c=\dfrac{z}{x}\)

Điều phải chứng minh tương đương với:

\(\dfrac{1}{2+\dfrac{x}{y}}+\dfrac{1}{2+\dfrac{y}{z}}+\dfrac{1}{2+\dfrac{z}{x}}\le1\\ \Leftrightarrow\dfrac{y}{2y+x}+\dfrac{z}{2z+y}+\dfrac{x}{2x+z}\le1\\ \Leftrightarrow\dfrac{2y}{2y+x}+\dfrac{2z}{2z+y}+\dfrac{2x}{2x+z}\le2\\ \Leftrightarrow\dfrac{x}{2y+x}+\dfrac{y}{2z+y}+\dfrac{z}{2x+z}\ge1\left(1\right)\)

Áp dụng bất đẳng thức bunhiacopxki dạng phân thức ta có:

\(\dfrac{x}{2y+x}+\dfrac{y}{2z+x}+\dfrac{z}{2x+z}=\dfrac{x^2}{x^2+2xy}+\dfrac{y^2}{y^2+2zx}+\dfrac{z^2}{z^2+2xy}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

=> bài toán được chứng minh

Dấu bằng xảy ra khi x=y=z=1 <=>a=b=c=1

ĐKXĐ : a;b;c  \(\ne0\)

Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2000}\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}-\dfrac{1}{a}\)

\(\Leftrightarrow\dfrac{b+c}{bc}=\dfrac{-\left(b+c\right)}{a\left(a+b+c\right)}\)

\(\Leftrightarrow\left(b+c\right)\left(\dfrac{1}{bc}+\dfrac{1}{a\left(a+b+c\right)}\right)=0\)

\(\Leftrightarrow\left(b+c\right).\dfrac{a\left(a+b+c\right)+bc}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(b+c\right).\dfrac{a^2+ab+ac+bc}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\dfrac{\left(b+c\right)\left(a+b\right)\left(a+c\right)}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b+c=0\\a+b=0\\a+c=0\end{matrix}\right.\left(1\right)\)

Từ (1) kết hợp a + b + c = 2000 ta được điều phải chứng minh

`1/a+1/b+1/c=1/(a+b+c)`

`<=>(a+b)/(ab)+(a+b)/(c(a+b+c))=0`

`<=>(a+b)(ab+ac+bc+c^2)=0`

`<=>(a+b)(a+c)(b+c)=0`

`=>` $\left[ \begin{array}{l}a=-b\\b=-c\\c=-a\end{array} \right.$

`=>` PT luôn tồn tại 2 số đối nhau

a: Gọi phân số cần tìm có dạng là \(\dfrac{a}{b}\left(b\ne0\right)\)

Theo đề, ta có: \(\dfrac{1}{3}< \dfrac{a}{b}< \dfrac{1}{2}\)

=>\(0,\left(3\right)< \dfrac{a}{b}< 0,5\)

=>\(\dfrac{a}{b}=0,4;\dfrac{a}{b}=0,42\)

=>\(\dfrac{a}{b}=\dfrac{2}{5};\dfrac{a}{b}=\dfrac{21}{25}\)

Vậy: Hai phân số cần tìm là \(\dfrac{2}{5};\dfrac{21}{25}\)

b: a/b<1

=>a<b

=>\(a\cdot c< b\cdot c\)

=>\(a\cdot c+ab< b\cdot c+ab\)

=>\(a\left(c+b\right)< b\left(a+c\right)\)

=>\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)

Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

=> \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2\left(\dfrac{c}{abc}+\dfrac{b}{abc}+\dfrac{a}{abc}\right)\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2.\dfrac{abc}{abc}\) = 4 ( vì a+b + c = abc)

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) => đpcm

Áp dụng bất đẳng thức Svacxo ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\)

Tương tự : \(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c};\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{a+2b}+\dfrac{3}{b+2c}+\dfrac{3}{c+2a}\)

Dấu = xảy ra khi a=b=c

\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\)(BĐT Cauchy Schawarz)(1)

tương tự \(=>\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c}\left(2\right)\)

\(=>\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\left(3\right)\)

(1)(2)(3)

\(=>3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)

\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\left(dpcm\right)\)

Chứng minh biểu thức \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\\ \Leftrightarrow\dfrac{a+2b}{a}+\dfrac{2\left(a+2b\right)}{b}\ge9\\ \Leftrightarrow\dfrac{2b}{a}+\dfrac{2a}{b}\ge4\\ \Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\left(cosi\right)\)

Đặt vế trái BĐT cần chứng minh là P

Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ( Tự chứng minh BĐT này ), ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\le\dfrac{1}{\dfrac{4}{a+b}}=\dfrac{a+b}{4}\left(1\right)\)

Tương tự: \(\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}\le\dfrac{b+c}{4}\left(2\right)\)

\(\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\le\dfrac{c+a}{4}\left(3\right)\)

Cộng \(\left(1\right),\left(2\right),\left(3\right)\) vế theo vế, ta được:

\(P\le\dfrac{a+b+b+c+c+a}{4}=\dfrac{a+b+c}{2}\)

Dấu ''='' xảy ra khi và chỉ khi a=b=c