Question

cho 3 số dương thỏa mãn a+b+c=3. Chứng minh rằng\(\frac{a\left(a+c-2b\right)}{1+ab}+\frac{b\left(b+a-2c\right)}{1+bc}+\frac{c\left(c+b-2a\right)}{1+ca}\ge0\)

( mình chọn chủ đề linh tinh nhá :V vì ko có )

Answer 1

\(\sum\frac{a\left(a+c-2b\right)}{1+ab}\ge0\Leftrightarrow\sum\frac{a\left(3-3b\right)}{1+ab}\ge0\Leftrightarrow\sum\frac{a\left(1-b\right)}{1+ab}\ge0\)

Ta có:

\(VT=\sum\frac{a\left(1-b\right)}{1+ab}=\sum\left(a-\frac{ab\left(1+a\right)}{1+ab}\right)\ge\sum\left(a-\frac{ab\left(1+a\right)}{2\sqrt{ab}}\right)\)

\(VT\ge\sum\left(a-\frac{1}{4}\left(2.1.\sqrt{ab}+2.a.\sqrt{ab}\right)\right)\ge\sum\left(a-\frac{1}{4}\left(1+ab+a^2+ab\right)\right)\)

\(\Rightarrow VT\ge3-\frac{3}{4}-\frac{1}{4}\left(a+b+c\right)^2=0\)

Dấu "=" xảy ra khi \(a=b=c=1\)