\(abc\ne0\)

\(abc\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)=abc\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\)

\(\Leftrightarrow a^2c+ab^2+bc^2=b^2c+ac^2+a^2b\)

\(\Leftrightarrow a^2c-b^2c+ab^2-a^2b+bc^2-ac^2=0\)

\(\Leftrightarrow c\left(a-b\right)\left(a+b\right)-ab\left(a-b\right)-c^2\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(ac+bc-ab-c^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(c\left(a-c\right)-b\left(a-c\right)\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(c-b\right)\left(a-c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\) (đpcm)

Ta có \(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=\frac{1}{a-b-c}\)

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

=> \(\frac{b-a}{ab}=\frac{a-b}{\left(a-b-c\right)c}\)

Khi b - a = 0

=> (b - a)(a - c)(b + c) = 0 (1)

Khi b - a \(\ne0\)

=> ab = -(a - b - c).c

=> ab = -ac + bc + c² 

=> ab + ac - bc - c² = 0

=> a(b + c) - c(b + c) = 0

=> (a - c)(b + c) = 0

=> (b - a)(a - c)(b + c) = 0 (2)

Từ (1)(2) => (b - a)(a - c)(b + c) = 0

=> b - a = 0 hoặc a - c = 0 hoặc b + c = 0

=> a = b hoặc a = c hoặc b = -c

Vậy tồn tại 2 số bằng nhau hoặc đối nhau

 a/b+b/c+c/a=b/a+c/b+a/c 
<=> a/b-b/a+b/c-c/b+c/a-a/c=0 
<=> a^2c-c^2a+c^2b-b^2c+b^2a-a^2b=0 
<=> ac(a-c)+bc(c-b)+ab(b-a)=0 
<=> ac(a-c)+bc(c-a+a-b)+ab(b-a)=0 
<=> ac(a-c)+bc(c-a)+bc(a-b)+ab(b-a)=0 
<=> (a-c)(a-b)c+(a-b)(c-a)b=0 
<=> (a-b)(c-a)(b-c)=0 
<=> a=b hay c=a hay b=c 
Vậy trong ba số a,b,c tồn tại 2 số =nhau

kết quả sẽ ra là

(a-b)(a-c)(b-c)=0

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

\(\frac{a^2c}{abc}+\frac{b^2a}{abc}+\frac{c^2a}{abc}=\frac{b^2c}{abc}+\frac{c^2a}{abc}+\frac{a^2b}{abc}\)

\(=>a^2c+b^2a+c^2a=b^2c+c^2a+a^2b\)

Vì \(c^2a=c^2a\)=> \(a^2c+b^2a=b^2c+a^2b\)

=>đpcm, hình như mình giải thiếu điều kiện thì phải 

ừ nhỉ, chỗ phần quy đồng

\(\frac{a^2c}{abc}+\frac{b^2a}{abc}+\frac{c^2b}{abc}=\frac{b^2c}{abc}+\frac{c^2a}{abc}+\frac{a^2b}{abc}\)

\(a^2c+b^2a+c^2b=b^2c+c^2a+a^2b\)

đến chỗ này tịt , bài nãy còn rút gọn được chứ phần này thì không

thôi, bạn suy nghĩ tiếp chỗ này nhé

Lời giải:

Ta có \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)

\(\Leftrightarrow \frac{ab^2+bc^2+ca^2}{abc}=\frac{a^2b+b^2c+c^2a}{abc}\)

\(\Leftrightarrow ab^2+bc^2+ca^2=a^2b+b^2c+c^2a\)

\(\Leftrightarrow ab^2+bc^2+ca^2-a^2b-b^2c-c^2a=0\)

\(\Leftrightarrow ab(b-a)+bc(c-b)+ac(a-c)=0\)

\(\Leftrightarrow ab(b-a)-bc[(b-a)+(a-c)]+ac(a-c)=0\)

\(\Leftrightarrow (b-a)(ab-bc)+(a-c)(ac-bc)=0\)

\(\Leftrightarrow b(b-a)(a-c)-c(a-c)(b-a)=0\)

\(\Leftrightarrow (b-a)(a-c)(b-c)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b=a\\a=c\\b=c\end{matrix}\right.\)

Do đó luôn tồn tại hai số bằng nhau (đpcm)

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

\(\Rightarrow\dfrac{a^2c}{abc}+\dfrac{b^2a}{abc}+\dfrac{c^2b}{abc}=\dfrac{b^2c}{abc}+\dfrac{a^2b}{abc}+\dfrac{c^2a}{abc}\)

\(\Rightarrow\dfrac{a^2c+b^2a+c^2b}{abc}=\dfrac{b^2c+a^2b+c^2a}{abc}\)

\(\Rightarrow a^2c+b^2a+c^2b=b^2c+a^2b+c^2a\)

\(\Rightarrow a^2c+b^2a+c^2b-b^2c-a^2b-c^2a=0\)

\(\Rightarrow\left(a^2c-c^2a\right)+\left(b^2a-a^2b\right)+\left(c^2b-b^2c\right)=0\)

\(\Rightarrow ac\left(a-c\right)+ab\left(b-a\right)+bc\left(c-b\right)=0\)

\(\Rightarrow ac\left(a-c\right)+ab\left(b-a\right)+bc\left(c-b+a-a\right)=0\)

\(\Rightarrow ac\left(a-c\right)+ab\left(b-a\right)+bc\left(c-a\right)+bc\left(a-b\right)\)

\(\Rightarrow c\left(a-c\right)\left(a-b\right)+b\left(a-b\right)\left(c-a\right)=0\)

\(\Rightarrow c\left(a-c\right)\left(a-b\right)-b\left(a-b\right)\left(a-c\right)=0\)

\(\Rightarrow\left(c-b\right)\left(a-c\right)\left(a-b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}c=b\\a=c\\a=b\end{matrix}\right.\)(Tồn tại ít nhất 2 số bằng nhau)

ĐKXĐ : \(a+b\ne0;a+c\ne0;b+c\ne0.\)

Từ \(\left(1\right)\Leftrightarrow\left(\frac{x-ab}{a+b}-c\right)+\left(\frac{x-ac}{a+c}-b\right)+\left(\frac{x-bc}{b+c}-a\right)=0\)

\(\Leftrightarrow\frac{x-ab-ac-bc}{a+b}+\frac{x-ac-ab-bc}{a+c}+\frac{a-bc-ab-ac}{b+c}=0\)

\(\Leftrightarrow\left(x-ab-bc-ca\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=0\)

\(\left(1\right)\) có vô số nghiệm \(\Leftrightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=0.\left(2\right)\)

Chẳng hạn ta chọn \(a=1,b=1.\)Để ( 2 ) xảy ra ta chọn c sao cho :

\(\frac{1}{2}+\frac{1}{1+c}+\frac{1}{1+c}=0\Leftrightarrow\frac{2}{1+c}=-\frac{1}{2}\Leftrightarrow c=-5.\)

Như vậy \(\left(1\right)\) có vô số nghiệm , chẳng hạn khi \(a=1,b=1,c=-5.\)

....................................................................................................................................................................................................................................

\(\frac{a^4c^3+b^4a^3+c^4b^3}{a^3b^3c^3}\)= \(\frac{b^4c+c^4a+a^4b}{abc}\)

\(\Rightarrow\)\(a^4c^3+b^4a^3+c^4b^3\)= \(b^4c+c^4a+a^4b\)

\(\Rightarrow\)\(a^4\left(c^3-b\right)+b^4\left(a^3-c\right)+c^4\left(b^3-a\right)\)= 0

suy ra c^3 -b = 0 hoặc a^3 -c = 0 hoặc b^3 -a = 0

suy ra   đpcm

đặt \(\hept{\begin{cases}x=\frac{a}{b^3}\\y=\frac{b}{c^3}\\z=\frac{c}{a^3}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{x}=\frac{b^3}{a}\\\frac{1}{y}=\frac{c^3}{b}\\\frac{1}{z}=\frac{a^3}{c}\end{cases}}\)khi đó  xyz=1

đề bài <=> x+y+z =1/x +1/y +1/z => x+y+z =yz+xz+xy

từ đó => xyz+  (x+y+z) -(xy+yz+xz)-1=0    <=> (x-1)(y-1)(z-1)=0

vây tồn tại x=1 =>a=b^3 (đpcm")

Bài 1 :

Áp dụng BĐT Cô - si cho 3 số không âm

\(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{a^3}{b^3}}+1\ge3\sqrt[3]{\sqrt{\frac{a^6}{b^6}}}=\frac{3a}{b}\)

\(\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{b^3}{c^3}}+1\ge3\sqrt[3]{\sqrt{\frac{b^6}{c^6}}}=\frac{3b}{c}\)

\(\sqrt{\frac{c^3}{a^3}}+\sqrt{\frac{c^3}{a^3}}+1\ge3\sqrt[3]{\sqrt{\frac{c^6}{a^6}}}=\frac{3c}{a}\)

Cộng theo vế , ta được :

\(2\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)+3\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

\(\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+3\)

\(\Rightarrow2\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

\(\Rightarrow\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

Vậy \(\Rightarrow\left(\sqrt{\frac{a^3}{b^3}}+\sqrt{\frac{b^3}{c^3}}+\sqrt{\frac{c^3}{a^3}}\right)\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(đpcm\right)\)