\(VT=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ba+bc}+\dfrac{c^2}{cb+ca}\)
- Áp dụng BĐT Bunhiacopxki ta có:
\(\left(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ba+bc}+\dfrac{c^2}{cb+ca}\right)\left[\left(ab+ac\right)+\left(ba+bc\right)+\left(cb+ca\right)\right]\ge\left(a+b+c\right)^2\)
\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\left(1\right)\)
Mặt khác ta có: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow VT\ge\dfrac{3\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=\dfrac{3}{2}\left(đpcm\right)\)
- Dấu "=" xảy ra \(\Leftrightarrow a=b=c>0\)
`a/(b+c) + b/(a+c) + c/(b+a)`
`= a^2/(ab+ac) + b^2/(ab+bc) + c^2/(bc+ca)`.
Có: `a^2 + b^2 + c^2 >= ab + bc + ca` và `(a+b+c)^2 = a^2 + b^2 + c^2 >= 2(ab+bc+ca) (` Bạn đọc tự chứng minh `)`.
Áp dụng bất đăng thức Svac - xơ:
`a^2/(ab+ac) + b^2/(ab+bc) + c^2/(bc+ca) >= (a+b+c)^2/(ab + bc + ca + ab + bc + ca) = (a+b+c)^2/(2(ab+bc+ca)) = (a^2 + b^2 + c^2 + 2(ab+bc+ca))/(2(ab+bc+ca)) >= (ab+bc+ca+2(ab+bc+ca))/(2(ab+bc+ca)) = (3(ab+bc+ca))/(2(ab+bc+ca)) = 3/2`
Dấu bằng xảy ra `<=> a/(ab+ac) = b/(ab+bc) = c/(bc+ca)` hay `a = b = c`.
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
