Ta có : a/b < c/d => ad < cb
=>ad +ab < bc+ab
=> a(d+b) < b(a+c)
=> a/b < a+c/d+b (1)
Ta có : a/b < c/d => ad<cb
=> ad + cd < cb +cd
=> d(a+c) < c(b+d)
=> c/d > a+c/b+d (2)
Từ (1) và (2) => a/b < a+c/b+d < c/d

Sửa đề: \(a;b;c\ge0\) (nếu không thì không có max đâu cu!)

Ta có: \(P=a\left(b-c\right)\le ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\)

Đẳng thức xảy ra khi \(a=b=\frac{1}{2};c=0\)

Vậy..

Có:\(a+b+c=0\Rightarrow c=-a-b\)

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

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

Cho \(B=\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\)

\(1\)\(\Rightarrow B.\dfrac{c}{a-b}=\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right).\dfrac{c}{a-b}\)

\(\Rightarrow B.\dfrac{c}{a-b}=1+\dfrac{c}{a-b}\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\)

\(\Rightarrow B.\dfrac{c}{a-b}=1+\dfrac{c}{a-b}.\dfrac{b^2-bc+ac-a^2}{ab}\)

\(\Rightarrow B.\dfrac{c}{a-b}=1+\dfrac{c}{a-b}.\dfrac{\left(a-b\right)\left(c-a-b\right)}{ab}\)

\(\Rightarrow B.\dfrac{c}{a-b}=1+\dfrac{2c^2}{ab}=1+\dfrac{2c^3}{abc}\)

\(2\ B.\dfrac{a}{b-c}=\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right).\dfrac{a}{b-c}\)

\(\Rightarrow B\dfrac{a}{b-c}=1+\dfrac{a}{b-c}\left(\dfrac{a-b}{c}+\dfrac{c-a}{b}\right)\)

\(\Rightarrow B.\dfrac{a}{b-c}=1+\dfrac{a}{b-c}\left(\dfrac{ab-b^2+c^2-ac}{bc}\right)\)

\(\Rightarrow B.\dfrac{a}{b-c}=1+\left(\dfrac{\left(b-c\right)\left(b+c\right)-a\left(b+c\right)}{bc}\right)\dfrac{a}{b-c}\)

\(\Rightarrow B.\dfrac{a}{b-c}=1+\left(\dfrac{\left(b+c\right)\left(b-c-a\right)}{bc}\right).\dfrac{a}{b-c}\)

\(\Rightarrow B.\dfrac{a}{b-c}=1+\dfrac{2a^3}{abc}\)

\(3\ \)\(B.\dfrac{b}{c-a}=1+\dfrac{2b^3}{abc}\)

\(\Rightarrow A=\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\)

\(\Rightarrow\left(B.\dfrac{c}{a-b}\right)+\left(B.\dfrac{a}{b-c}\right)+\left(B.\dfrac{b}{c-a}\right)\)

\(\Rightarrow1+\dfrac{2a^3}{abc}+1+\dfrac{2b^3}{abc}+1+\dfrac{2c^3}{abc}\)

\(\Rightarrow3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}\)

Áp dụng hằng đẳng thức mở rộng \(a^3+b^3+c^3=3abc\) khi \(a+b+c=0\)

\(\Rightarrow A=9\)

nhailaier you ngáo ak

Cho \(A=a+b+c=0\)

Tính \(A=\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\) chắc chắn bằng 0 rồi :V

Áp dụng BĐT Cosi dạng engel cho 3 số dương ta có:

\(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)

Dấu "=" xảy ra khi \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

Ta thấy \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\)đều là số dương

Vì thế nên ta sẽ áp dụng bđt cô-si dạng engel:

\(\frac{x^2+y^2+z^2}{a+b+c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)

Vậy đẳng thức chỉ xảy ra khi \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

Giải cách khác được ko bn ơi mình chưa học BĐT Cô-si