\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)=> \(a+b+c=\frac{ab+bc+ac}{abc}=ab+bc+ac\)

Ta có \(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(abc-1\right)+a+b+c-ab-bc-ac=0\)

=> có ít nhất 1 trong 3 số a,b,c bằng 1

Vậy có ít nhất 1 trong 3 số a,b,c bằng 1

Thay 1 = abc ta có: \(a+b+c=\frac{abc}{a}+\frac{abc}{b}+\frac{abc}{c}\)

<=> a + b + c = bc + ac + ab

<=> (a - ac) + (b - bc) + (c - ab) = 0 

<=> a(1 - c) + b(1 - c) + (c - \(\frac{1}{c}\)) = 0 

<=> ca(1 - c) + cb(1 - c) + (c - 1)(c + 1) = 0 

<=> (1 - c)(ca + cb - c - 1) = 0 

<=> (1 - c)[c(a -1) + (cb - abc)]= 0 

<=> (1 - c)[c(a - 1) + cb(1 - a)]= 0 

<=> (1 - c)(a - 1)(c - cb) = 0

<=> (1 - c)(a - 1)(1 - b).c = 0 <=> a = 1 hoặc b = 1 hoặc c = 1

Vậy.... 

http://olm.vn/hoi-dap/question/179947.html

Cso bài mk đó,lik-e nha

2) M = (x²⁵ + 1 + 1 + 1 + 1) - 5x⁵ + 2

Áp dụng BĐT Cô - si cho 5 số dương x²⁵; 1;1;1;1 ta có: x²⁵ + 1 + 1 + 1 + 1 \(\ge\)5.\(\sqrt[5]{x^{25}.1.1.1.1}=x^5\) = 5x⁵

=> M \(\ge\) 5x⁵ - 5x⁵ + 2 = 2

Vậy M nhỏ nhất = 2 khi x²⁵ = 1 => x = 1

\(ab=\frac{1}{c};c=\frac{1}{ab}\)

\(a+b+c-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a+b+\frac{1}{ab}-\frac{1}{a}-\frac{1}{b}-ab\)

\(=\left(a+b-ab-1\right)+\left(\frac{1}{ab}-\frac{1}{a}-\frac{1}{b}+1\right)\)

\(=-\left(a-1\right)\left(b-1\right)+\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\)

\(=-\left(a-1\right)\left(b-1\right)+\frac{\left(a-1\right)\left(b-1\right)}{ab}\)

\(=-\left(a-1\right)\left(b-1\right)+\left(a-1\right)\left(b-1\right)c\)

\(=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)

Do biểu thức ban đầu dương nên ta có đpcm

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

2 ) Ta có :

\(\frac{a}{b}>\frac{a}{b+c+a}\) 

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

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

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

biến đổi tương đương đưa về (a-1)(b-1)(c-1)=0

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

\(\Leftrightarrow a+b+c=\frac{ab+bc+ac}{abc}\)

\(\Leftrightarrow a+b+c=ab+bc+ac\left(abc=1\right)\)

\(\Leftrightarrow1+a+b+c-ab-bc-ac-1=0\)

\(\Leftrightarrow abc+a+b+c-ab-bc-ac-1=0\)

\(\Leftrightarrow ab\left(c-1\right)-a\left(c-1\right)-b\left(c-1\right)+c-1=0\)

\(\Leftrightarrow\left(ab-a-b+1\right)\left(c-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\)

\(\Leftrightarrow\)a = 1 hoặc b = 1 hoặc c = 1

=> Đpcm 

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\ge\frac{9}{6}=\frac{3}{2}\)

\(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\ge\frac{9}{ab+bc+ca}\ge\frac{27}{\left(a+b+c\right)^2}=\frac{27}{36}=\frac{3}{4}\)

\(\frac{1}{abc}\ge\frac{1}{\left(\frac{a+b+c}{3}\right)^3}=\frac{27}{\left(a+b+c\right)^3}\ge\frac{27}{6^3}=\frac{1}{8}\)

Cộng lại ta được:

\(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\ge\frac{27}{8}\left(đpcm\right)\)

Dấu "=" xảy ra tại \(a=b=c=2\)