Đặt \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}=k\)

=> \(\begin{cases}a+b=k.\left(c+d\right)=k.c+k.d\\a-2b-k.\left(c-2d\right)=k.c-k.2d\end{cases}\)

=> (a + b) - (a - 2b) = (k.c + k.d) - (k.c - k.2d)

=> a + b - a + 2b - k.c + k.d - k.c + k.2d

=> 3b = 3kd

=> b = kd

Mà a + b = k.c + k.d

=> a = k.c

=> \(\frac{a}{b}=\frac{k.c}{k.d}=\frac{c}{d}\left(đpcm\right)\)

Cách 2:

Ta có: \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)

=> (a + b).(c - 2d) = (c + d).(a - 2b)

=> (a + b).c - (a + b).2d = (c + d).a - (c + d).2b

=> ac + bc - 2ad - 2bd = ac + ad - 2bc - 2bd

=> ac + bc - 2ad - 2bd - ac - ad + 2bc + 2bd = 0

=> 3bc - 3ad = 0

=> 3bc = 3ad

=> bc = ad

=> \(\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

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

=> a = b = c = d

=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)

D = 1 + 1 + 1 + 1 = 4

Đặt cái ban đầu là A

Dầu tiên ta có

\(\text{(3a+c)(a+2b+c)+(3b+d)(b+2c+d)+(3c+a)(c+2d+a)+(3d+b)(d+2a+b)}\)

\(=4\left(a+b+c+d\right)^2\)

Ta có: \(\frac{a-b}{a+2b+c}+\frac{1}{2}=\frac{1}{2}.\frac{3a+c}{a+2b+c}=\frac{1}{2}.\frac{\left(3a+c\right)^2}{\left(3a+c\right)\left(a+2b+c\right)}\)

Tương tự ta có

\(\frac{b-c}{b+2c+d}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3b+d\right)^2}{\left(3b+d\right)\left(b+2c+d\right)}\)

\(\frac{c-d}{c+2d+a}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3c+a\right)^2}{\left(3c+a\right)\left(c+2d+a\right)}\)

\(\frac{d-a}{d+2a+b}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3d+b\right)^2}{\left(3d+b\right)\left(d+2a+b\right)}\)

Cộng vế theo vế ta được

\(\frac{a-b}{a+2b+c}+\frac{1}{2}+\frac{b-c}{b+2c+d}+\frac{1}{2}+\frac{c-d}{c+2d+a}+\frac{1}{2}+\frac{d-a}{d+2a+b}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3d+b\right)^2}{\left(3d+b\right)\left(d+2a+b\right)}+\frac{1}{2}.\frac{\left(3c+a\right)^2}{\left(3c+a\right)\left(c+2d+a\right)}+\frac{1}{2}.\frac{\left(3b+d\right)^2}{\left(3b+d\right)\left(b+2c+d\right)}+\frac{1}{2}.\frac{\left(3a+c\right)^2}{\left(3a+c\right)\left(a+2b+c\right)}\)

\(\ge\frac{1}{2}.\frac{\left(3a+c+3b+d+3c+a+3d+b\right)^2}{\left(3a+c\right)\left(a+2b+c\right)+\left(3b+d\right)\left(b+2c+d\right)+\left(3c+a\right)\left(c+2d+a\right)+\left(3d+b\right)\left(d+2a+b\right)}\)

\(=\frac{1}{2}.\frac{16\left(a+b+c+d\right)^2}{4\left(a+b+c+d\right)^2}=2\)

\(\Rightarrow A+2\ge2\)

\(\Leftrightarrow A\ge0\)

=4(a+b+c+d)2

Ta có: a−ba+2b+c +12 =12 .3a+ca+2b+c =12 .(3a+c)2(3a+c)(a+2b+c) 

Tương tự ta có

b−cb+2c+d +12 =12 .(3b+d)2(3b+d)(b+2c+d) 

c−dc+2d+a +12 =12 .(3c+a)2(3c+a)(c+2d+a) 

d−ad+2a+b +12 =12 .(3d+b)2(3d+b)(d+2a+b) 

Cộng vế theo vế ta được

a−ba+2b+c +12 +b−cb+2c+d +12 +c−dc+2d+a +12 +d−ad+2a+b +12 =12 .(3d+b)2(3d+b)(d+2a+b) +12 .(3c+a)2(3c+a)(c+2d+a) +12 .(3b+d)2(3b+d)(b+2c+d) +12 .(3a+c)2(3a+c)(a+2b+c) 

≥12 .(3a+c+3b+d+3c+a+3d+b)2(3a+c)(a+2b+c)+(3b+d)(b+2c+d)+(3c+a)(c+2d+a)+(3d+b)(d+2a+b) 

=12 .16(a+b+c+d)24(a+b+c+d)2 =2

⇒A+2≥2

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{4a}{4c}=\frac{2b}{2d}=\frac{4a-2b}{4c-2d}=\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a+2b}{5c+2d}\)

Suy ra \(\frac{4a-2b}{4c-2d}=\frac{5a+2b}{5c+2d}\)Suy ra điều phải chứng minh: \(\frac{4a-2b}{5a+2b}=\frac{4c-2d}{5c+2d}\)

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