Áp dụng BĐT AM-GM ta có:
\(\dfrac{a}{2b+2c-a}=\dfrac{3a^2}{3a\left(2b+2c-a\right)}\ge\dfrac{3a^2}{\dfrac{\left(3a+2b+2c-a\right)^2}{4}}\)
\(\dfrac{12a^2}{\left(3a+2b+2c-a\right)^2}\)\(=\dfrac{12a^2}{\left(2a+2b+2c\right)^2}\)
Tương tự ta cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b}{2a+2c-b}\ge\dfrac{12b^2}{\left(2a+2b+2c\right)^2};\dfrac{c}{2a+2b-c}\ge\dfrac{12c^2}{\left(2a+2b+2c\right)^2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{12\left(a^2+b^2+c^2\right)}{4\left(a+b+c\right)^2}\ge\dfrac{4\left(a+b+c\right)^2}{4\left(a+b+c\right)^2}=1\)
Đẳng thức xảy ra khi \(a=b=c\)
\(\dfrac{a}{2b+2c-a}+\dfrac{b}{2c+2a-b}+\dfrac{c}{2a+2b-c}\)
\(=\dfrac{a^2}{2ab+2ac-a^2}+\dfrac{b^2}{2bc+2ba-b^2}+\dfrac{c^2}{2ca+2cb-c^2}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)-a^2-b^2-c^2}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+a^2+b^2+c^2-a^2-b^2-c^2}=1\)
Dấu = xảy ra khi a = b = c
