\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)

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

   \(\dfrac{a}{c}\)  =  \(\dfrac{5a}{5c}\) = \(\dfrac{3b}{3d}\) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

      \(\dfrac{a}{c}\) =   \(\dfrac{5a+3b}{5c+3d}\) (1) 

       \(\dfrac{a}{c}\) = \(\dfrac{5a-3b}{5c-3d}\)  (2)

Kết hợp (1) và (2) ta có:

       \(\dfrac{5a+3b}{5c+3d}\) =  \(\dfrac{5a-3b}{5c-3d}\) 

⇒   \(\dfrac{5a+3b}{5a-3b}\) =  \(\dfrac{5c+3d}{5c-3d}\) (đpcm)

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

      \(\dfrac{a}{b}\) =  \(\dfrac{3a}{3b}\) = \(\dfrac{2c}{2d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

     \(\dfrac{a}{b}\) = \(\dfrac{3a+2c}{3b+2d}\) (đpcm)

Có \(\dfrac{a}{b}=\dfrac{c}{d}< =>ad=bc\)

Xét \(\dfrac{a+c}{b+d}-\dfrac{a-c}{b-d}\)

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

= \(\dfrac{ab-ad+bc-cd-ab+bc-da+cd}{\left(b+d\right)\left(b-d\right)}\) 

= 0

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

lol

\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\\ \Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{2b}{2d}=\dfrac{b}{d}\left(1\right)\\ \dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{2a}{2c}=\dfrac{a}{c}\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

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

Làm tương tự với 3 phân số còn lại và cộng vế với vế

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

Làm tương tự với 3 phân số còn lại và cộng vế với vế

\(VT=\dfrac{\left(a+c\right)^2}{\left(a+c\right)\left(a+b\right)}+\dfrac{\left(b+d\right)^2}{\left(b+c\right)\left(b+d\right)}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)\left(c+d\right)}+\dfrac{\left(d+b\right)^2}{\left(d+a\right)\left(d+b\right)}\)

\(VT\ge\dfrac{\left(2a+2b+2c+2d\right)^2}{\left(a+b\right)\left(a+c\right)+\left(b+c\right)\left(b+d\right)+\left(a+c\right)\left(c+d\right)+\left(a+d\right)\left(b+d\right)}=\dfrac{4\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=4\)

Dấu "=" xảy ra khi \(a=b=c=d\)

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a+b}{a}=\dfrac{bk+b}{bk}=\dfrac{k+1}{k}\)

\(\dfrac{c+d}{c}=\dfrac{dk+d}{d}=\dfrac{k+1}{k}\)

=>(a+b)/a=(c+d)/c

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

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

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

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

Áp dụng cauchy-schwarz:

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+e}+\dfrac{d}{e+a}+\dfrac{e}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{cd+ce}+\dfrac{d^2}{ed+ad}+\dfrac{e^2}{ae+be}\ge\dfrac{\left(a+b+c+d\right)^2}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\)

Giờ chỉ cần chứng minh

\(ab+ac+ad+ae+bc+bd+be+cd+ce+de\le\dfrac{2}{5}\left(a+b+c+d+e\right)^2\)

\(\Leftrightarrow ab+ac+ad+ae+bc+bd+be+cd+ce+de\le2\left(a^2+b^2+c^2+d^2+e^2\right)\)

điều này hiển nhiên đúng theo AM-GM:

\(ab\le\dfrac{a^2+b^2}{2};ac\le\dfrac{a^2+c^2}{2};ad\le\dfrac{a^2+d^2}{2}...\)

Cứ vậy ta thu được đpcm .Dấu = xảy ra khi a=b=c=d=e

P/s: : ]

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)

\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)

Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)

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

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

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