Giả sử c là số ở giửa a và b. khi đó \(\left(b-c\right)\left(c-a\right)\ge0\)
Ta chứng minh :
\(VT\le c\left(\dfrac{b^2}{2b^2+a^2+c^2}+\dfrac{a^2}{2a^2+b^2+c^2}\right)+\dfrac{abc}{a^2+b^2+2c^2}\)(*)
\(\Leftrightarrow\dfrac{\left(c-a\right)\left(b-c\right)\left(b^2+c^2-bc+a^2\right)}{\left(a^2+c^2+2b^2\right)\left(b^2+a^2+2c^2\right)}\ge0\) (Đúng)
Áp dụng BĐT AM-GM:
\(VT\le\dfrac{c}{4}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{a^2}{a^2+b^2}+\dfrac{a^2}{a^2+c^2}\right)+\dfrac{abc}{2ac+2bc}\)
\(\le\dfrac{c}{4}\left(1+\dfrac{b^2}{2bc}+\dfrac{a^2}{2ac}\right)+\dfrac{\dfrac{\left(a+b\right)^2}{4}}{2\left(a+b\right)}=\dfrac{c}{4}+\dfrac{a+b}{8}+\dfrac{a+b}{8}\)
\(=\dfrac{a+b+c}{4}\)( \(ĐpcM\))
Dấu = xảy ra khi a=b=c
