Đề bài có nhầm lẫn gì ko nhỉ?

\(T=\dfrac{ab}{a^2+b^2+ab}+\dfrac{bc}{b^2+c^2+2bc}+\dfrac{ca}{c^2+a^2+ca}\le\dfrac{ab}{2ab+ab}+\dfrac{bc}{2bc+bc}+\dfrac{ca}{2ca+ca}=1\)

\(VT=\dfrac{1}{\dfrac{a}{b}+\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{a}{b}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{b}{c}+1}\)

\(\left(\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\right)\rightarrow\left(x^3,y^3,z^3\right)\)\(\Rightarrow xyz=1\).

\(VT=\sum\dfrac{1}{x^3+y^3+1}\le\sum\dfrac{1}{xy\left(x+y\right)+xyz}=\sum\dfrac{z}{x+y+z}=1\)

Dấu = xảy ra khi x=y=z=1 hay a=b=c

wtf

Cauchy-SChwarz:

\(VT=\sum_{cyc}\frac{ab}{a^2+ab+bc}\le\frac{\sum_{cyc}\left(a^2b^2+ab^2c+abc^2\right)}{\left(ab+bc+ca\right)^2}=\frac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=1\)

Dau "=" khi a=b=c\(\in R^+\)

còn tick nữa tui đủ 145 mà ai kiết zợ

bài này tôi có thể làm đc nhưng có điều bạn phải tick cho tối đa

`Answer:`

Thêm điều kiện `a,b,c\ne0` nhé.

\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)

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

\(\Rightarrow\hept{\begin{cases}\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}\\\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\\\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{a}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{a}\\\frac{1}{c}=\frac{1}{b}\end{cases}}\)

\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

`=>a=b=c`

Lúc này `M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{1.1+1.1+1.1}{1^2+1^2+1^2}=3/3=1`

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

=> \(\dfrac{abc}{ac+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)

=> ac + bc = ab + ac = bc + ab (do abc \(\ne0\))

=> ac + bc - ab - ac = 0

=> bc - ab = 0

=> b(c - a) = 0

Mà b \(\ne0\) nên c - a = 0 => c = a

Tương tự ta có: a = b

Từ đó có: a = b = c

Thay vào M được:

\(M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

Do \(a,b,c\ne0\)

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)

\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{a}{ac}+\dfrac{c}{ac}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\b=a\end{matrix}\right.\) \(\Rightarrow a=b=c\)

\(\Rightarrow M=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)

\(\dfrac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)+\left(ab+bc+ac\right)^2}{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}\)

\(=\dfrac{a^3+a^2b+a^2c+ab^2+b^2+b^2c+ac^2+bc^2+c^3+a^2b^2+b^2c^2+a^2c^2+2ab^2c+2a^2bc+2abc^2}{a^2+b^2+c^2+2ab+2bc+2ac-\left(ab+bc+ac\right)}\)

\(=\dfrac{a^3+a^2b+a^2c+ab^2+b^3+b^2c+ac^2+bc^2+c^3+a^2b^2+b^2c^2+a^2c^2-2ab^2c+2a^2bc+2abc^2}{a^2+b^2+c^2+ab+ac+bc}\)