a)  đặt \(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow a=b.k;c=d.k\)

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

    b)          bó tay

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

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

\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)

\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)

*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)

Ta có: - a/b=c/d=2c/2d => a/b=2c/2d

Áp dụng tỉ lệ thức ta có:

a/b=2c/2d=(a+2c)/(b+2d) (1)

         - a/b=c/d=3c/3d =>a/b=3c/3d

Áp dụng tỉ lệ thức ta có:

a/b=3c/3d=(a-3c)/(b-3d) (2)

Từ (1) và (2) =>(a+2c)/(b+2d)=(a-2c)/(b-2d)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Khi đó : \(\frac{a+2c}{b+2d}=\frac{bk+2dk}{b+2d}=\frac{k\left(b+2d\right)}{b+2d}=k\left(1\right)\)

\(\frac{a-3c}{b-3d}=\frac{bk-3dk}{b-3d}=\frac{k\left(b-3d\right)}{b-3d}=k\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\frac{a+2c}{b+2d}=\frac{a-3c}{b-3d}\left(\text{đpcm}\right)\)