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

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

Làm lại lun ._.

Nguyễn Huy Tú chắc làm sai rồi

Chứng minh:

Ta có: \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)

\(\Rightarrow\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}\)

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

\(\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}=\dfrac{2a+13b+3a-7b}{2c+13d+3c-7d}=\dfrac{5a+6b}{5c+6d}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left\{{}\begin{matrix}a=b\\c=d\end{matrix}\right.\Rightarrow\dfrac{a}{a}=\dfrac{c}{c}\)

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

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

Sai !!!! TC DTSBN ko có điều ngược lại !!!!

Áp dụng tính chất của dãy tỉ số bằng nhau:

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

\(\dfrac{a}{3b}=\dfrac{1}{3}\Rightarrow a=b\) __( 1 )__

\(\dfrac{b}{3c}=\dfrac{1}{3}\Rightarrow b=c\) __( 2 )__

\(\dfrac{c}{3d}=\dfrac{1}{3}\Rightarrow c=d\) __( 3 )__

\(\dfrac{d}{3a}=\dfrac{1}{3}\Rightarrow d=a\) __ ( 4 )__

Từ ( 1 ), ( 2 ), ( 3 ), ( 4 ) suy ra: \(a=b=c=d\)

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{3k+5}{3k-5}\)

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

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

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)

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

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

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

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

Áp dụng bđt AM-GM:

\(\dfrac{a^3b}{c}+\dfrac{b^3c}{a}+\dfrac{c^3a}{b}+\dfrac{a^3c}{b}+\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge6\sqrt[6]{\dfrac{a^8b^8c^8}{a^2b^2c^2}}=6\sqrt[6]{a^6b^6c^6}=6abc\)Dấu "=" xảy ra khi \(a=b=c\)

Ta có: \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)

\(\Leftrightarrow\dfrac{2a+13b}{2c+13d}=\dfrac{3a-7b}{3c-7d}\)

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

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

hay \(\dfrac{a}{b}=\dfrac{c}{d}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

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

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

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

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)