Áp dụng tính chất DTS bằng nhau: 

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

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

Vậy....

Tự tl v!

Áp dụng tính chất DTS bằng nhau ,ta có: 

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

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

Vậy....

con này max dễ

1.

\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)

\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)

Dấu "=" khi \(a=b=c\)

2.

\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)

Dấu "=" khi \(a=b=c=d\)

Thục Trinh, tran nguyen bao quan, Phùng Tuệ Minh, Ribi Nkok Ngok, Lê Nguyễn Ngọc Nhi, Tạ Thị Diễm Quỳnh,

Nguyễn Huy Thắng, ?Amanda?, saint suppapong udomkaewkanjana

Help me!

Bài thứ hai đó áp dụng bđt cauchy showas là ra rồi sử dụng tch bắc cầu tệ.

a) \(\hept{\begin{cases}\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\\\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{2b}{2d}=\frac{3a+2b}{3c+2d}\end{cases}}\)

\(\Rightarrow\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)

\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)

b) Chứng minh tương tự 

ko biet nghen

Áp dụng BĐT Cauchy-Schwarz ta có:

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

\(\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\). Mà theo BĐT AM-GM ta có:

\(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}=\frac{\left(a+b+c+d\right)^2}{2\left[\left(a+b\right)\left(c+d\right)+\left(a+c\right)\left(b+d\right)+\left(a+d\right)\left(b+c\right)\right]}\ge\frac{2}{3}\)

Đẳng thức xảy ra khi a=b=c=d

là \(\frac{2}{3}\) nha

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

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

Do đó :

       \(\frac{a}{3b}=\frac{1}{3}\Rightarrow\frac{a}{b}.\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{a}{b}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow a=b\)

       \(\frac{b}{3c}=\frac{1}{3}\Rightarrow\frac{b}{c}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{b}{c}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow b=c\)

       \(\frac{c}{3d}=\frac{1}{3}\Rightarrow\frac{c}{d}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{c}{d}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow c=d\)

       \(\frac{d}{3a}=\frac{1}{3}\Rightarrow\frac{d}{a}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{d}{a}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow d=a\)

\(\Rightarrow a=b=c=d\)