Ta có
\(A=\left(a-b+c\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)=3+\left(\frac{a}{c}+\frac{c}{a}\right)-\left(\frac{a}{b}+\frac{b}{a}\right)-\left(\frac{b}{c}+\frac{c}{b}\right)\)
áp dụng bđt Cauchy ta có
\(A\ge3+2-2-2=1\)(đpcm)
Dấu "=" xảy ra khi a=b=c=1
\(\left(a-b+c\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge1\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c+a\right)\ge0\)(đúng)
Vậy bài toán được chứng minh
em làm chi tiết bài anh ali nha!
BĐT \(\Leftrightarrow\left(a-b\right)\left(\frac{1}{a}-\frac{1}{b}\right)+\frac{\left(a-b\right)}{c}+c\left(\frac{1}{a}-\frac{1}{b}\right)+1\ge1\)
\(\Leftrightarrow\left(a-b\right)\left(\frac{1}{a}-\frac{1}{b}\right)+\frac{\left(a-b\right)}{c}+c\left(\frac{1}{a}-\frac{1}{b}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\frac{b-a}{ab}\right)+\frac{\left(a-b\right)}{c}+c\left(\frac{b-a}{ab}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)-c\left(\frac{a-b}{ab}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}-\frac{c}{ab}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left[\left(\frac{1}{a}-\frac{c}{ab}\right)+\left(\frac{1}{c}-\frac{1}{b}\right)\right]\ge0\)
\(\Leftrightarrow\left(a-b\right)\left[\frac{b-c}{ab}+\frac{b-c}{bc}\right]\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)\left(b-c\right)}{b}\left(\frac{1}{a}+\frac{1}{c}\right)=\frac{\left(a-b\right)\left(b-c\right)\left(c+a\right)}{abc}\ge0\)
BĐT tới đây đúng, ta có đpcm.
