Ta sẽ chứng minh:
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{3}{2}\)
Thật vậy,ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\)
Áp dụng bất đẳng thức Cauchy-Schwarz: \(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\) (1)
Ta sẽ chứng minh:
\(\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{3}{2}\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{ab+bc+ac}\ge3\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{3}\ge ab+bc+ac\) *đúng*
\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{3}{2}\) (2)
Từ (1) và (2) ta có: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{3}{2}\)
Hay \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c>0\)
F.C Kien Nguyen Nguyễn ☠Văn☠Xuân☠ Nguyễn Thị Diễm Quỳnh Nam Nguyễn Unruly KidNguyễn Thanh Hằngkiệt đẹp trai @^-^ Chúa tể hắc ám giúp mk với :((
