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

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

\(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}=\dfrac{2}{b}\)

Tương tự:

\(\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{2}{a}\) ; \(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{2}{c}\)

Cộng vế:

\(2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)

\(\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

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

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

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

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

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

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)

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

Vậy...

Hằng đẳng thức:

\(\left(x-y-z\right)^2=x^2+y^2+z^2+2\left(yz-xy-zx\right)=x^2+y^2+z^2-2\left(xy+xz-yz\right)\)

\(\Rightarrow x^2+y^2+z^2=\left(x-y-z\right)^2+2\left(xy+xz-yz\right)\)

Giờ thay \(x=\dfrac{1}{a}\) ; \(y=\dfrac{1}{b}\); \(z=\dfrac{1}{c}\) là ra cái người ta làm

C. \(\dfrac{b}{d}=\dfrac{c}{a}\)

Chúc bạn học tốt!!

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\) (đpcm)