Ta có : \(\dfrac{ab+1}{b}=\dfrac{ac+1}{a}=\dfrac{bc+1}{c}=\dfrac{ab+1+ac+1-bc-1}{b+a-c}=\dfrac{ab+ac-bc+1}{b+a-c}\)
=> \(\dfrac{ab}{b}=\dfrac{ac}{a}=\dfrac{bc}{c}=1\Rightarrow\dfrac{a}{b}=\dfrac{c}{a}=\dfrac{b}{c}=1\)
Điều này chỉ xảy ra khí a=b=c ( đpcm)
Cho a,b,c thỏa mãn abc=1
Tính B= \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
Cho a + b + c = 1 (a,b,c khác 1,2). Chứng minh
\(\dfrac{c+ab}{a^2+b^2+abc-1}+\dfrac{a+bc}{b^2+c^2+abc-1}+\dfrac{b+ac}{a^2+c^2+abc-1}=\dfrac{bc+ac+ab+8}{\left(a-2\right)\left(b-2\right)\left(a-2\right)}\)
Cho a,b,c là các số thực dương thỏa mãn a+b+c=1
Chứng minh rằng A=\(\dfrac{ab+c}{c+1}+\dfrac{bc+a}{a+1}+\dfrac{ac+b}{b+1}\le1\)
Cho a, b, c >0 thỏa mãn: abc=1. CM: \(\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{b^2-bc+c^2}+\dfrac{1}{c^2-ac+a^2}\le a+b+c\)
chứng minh rằng:\(\dfrac{a+b}{ab+c^2}+\dfrac{b+c}{bc+a^2}+\dfrac{c+a}{ac+b^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Cho 3 số a;b;c thỏa mãn a.b.c=1.CMR :\(\dfrac{a}{ab+a+1}\)+\(\dfrac{b}{bc+b+1}\)+\(\dfrac{c}{ac+c+1}\)=1
Cho abc=1. CMR \(\dfrac{a}{ab}+a+1+\dfrac{b}{bc}+b+1+\dfrac{c}{ac}+c+1=1\)
Cho abc=1.CMR:
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+}=1\)
Cho a,b,c>0 thỏa mãn ab+bc+ac=3. Cmr:
\(P=\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}\le1\)
