\(\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}\ge\frac{a-d}{a+b}\)

\(\Leftrightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}-\frac{a-d}{a+b}\ge0\)

\(\Leftrightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}+\frac{d-a}{a+b}\ge0\)

\(\Leftrightarrow\left(\frac{a-b}{b+c}+1\right)+\left(\frac{b-c}{c+d}+1\right)+\left(\frac{c-d}{a+d}+1\right)+\left(\frac{d-a}{a+b}+1\right)\ge4\)

\(\Leftrightarrow\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{a+d}+\frac{d+b}{a+b}\ge4\)

\(\Leftrightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge4\)(1)

Áp dụng BĐT AM-GM ta có:

\(\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{a+d}+\frac{d+b}{a+b}\ge\)\(\left(a+c\right)\frac{2}{\sqrt{\left(b+c\right)\left(a+d\right)}}+\left(b+d\right)\frac{2}{\sqrt{\left(c+d\right)\left(a+b\right)}}\ge\frac{4\left(a+c\right)}{a+b+c+d}+\frac{4\left(b+d\right)}{a+b+c+d}=\frac{4\left(a+b+c+d\right)}{a+b+c+d}=4 \left(2\right)\)Từ (1) và (2) \(\Rightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}\ge\frac{a-d}{a+b}\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{1}{b+c}=\frac{1}{a+d}\\\frac{1}{c+d}=\frac{1}{a+b}\end{cases}}\Leftrightarrow\hept{\begin{cases}b+c=a+d\\c+d=a+b\end{cases}}\Leftrightarrow a=b=c=d\)

vì sao                                          

 (a+c)(2/căn bậc 2 của(b+c)(a+d))+(b+d)(2/căn bậc 2 của (c+d)(a+b))
>=(4(a+c)/a+b+c+d) +4(b+d)/a+b+c+d

(căn bậc 2 máy mink ko viết đc)

\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)

Có:

\(\Rightarrow\frac{a\left(b+d\right)}{b\left(b+d\right)}< \frac{b\left(a+c\right)}{b\left(b+d\right)}\)

\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)

\(\Rightarrow\frac{d\left(a+c\right)}{d\left(b+d\right)}< \frac{c\left(b+d\right)}{d\left(b+d\right)}\)

\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

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

\(\left(a+d\right)^2-\left(b+c\right)^2=\left(a-d\right)^2-\left(b-c\right)^2\)

\(\left(a+d\right)^2-\left(a-d\right)^2=\left(b+c\right)^2-\left(b-c\right)^2\)

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

\(2d\times2a=2b\times2c\)

\(ad=bc\)

\(\frac{a}{c}=\frac{b}{d}\left(\text{đ}pcm\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)

Ta có:\(\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)

\(\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)

Từ (1) và (2), ta có: \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

_Học tốt_

Thanks

Ta có :

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}=\left(\frac{a+b+c}{b+c+d}\right)^3\) (1)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\) (2)

Từ (1) và (2) suy ra điều phải chứng minh

Ta có : a+b/b+c = c+d/d+a 
=> (a+b)/(c+d)= (b+c)/(d+a) 
=> (a+b)/(c+d)+1=(b+c)/(d+a)+1 
hay: (a+b+c+d)/(c+d)=(b+c+d+a)/(d+a) 
- Nếu a+b+c+d khác 0 thì : c+d=d+a => c=a (1)
- Nếu a+b+c+d = 0 (2) 

Từ (1) và (2) 

\(\RightarrowĐPCM\)

Ta có : \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

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

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

Hoặc \(\frac{a+b+c+d}{c+d}=\frac{b+c+d+a}{d+a}\)

Nếu a + b + c + d khác 0 thì c + d = d + a => c = a ( hoặc a = c )

Nếu a + b + c + d = 0 ( đpcm )

Ta có:\(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

\(\implies\)\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)

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

\(\implies\) \(\frac{a+b+c+d}{c+d}=\frac{a+b+c+d}{d+a}\)

\(\implies\) \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)

\(\implies\) \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)

\(\implies\)\(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}=\frac{1}{d+a}\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c+d=d+a\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c=a\end{cases}}\)