đặt a = 2x + y + z; b = 2y + z + x; c = 2z + x + y (a; b ; c > 0)
=> a + b + c = 4.(x+ y + z) => x + y + z = (a+ b+ c) / 4
=> x = a - (x+ y + z) = a - (a+ b + c) / 4
y = b - (x + y + z) = b - (a+b+c) / 4
z = c - (x+y + z) = c - (a+b+c)/ 4
Khi đó : \(VT=1-\frac{a+b+c}{4a}+1-\frac{a+b+c}{4b}+1-\frac{a+b+c}{4c}\)
\(VT=3-\left(\frac{a+b+c}{4a}+\frac{a+b+c}{4b}+\frac{a+b+c}{4c}\right)=3-\frac{1}{4}.\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(VT=3-\frac{1}{4}.\left(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\right)=3-\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\)
Với a, b > 0 ta có: a/b + b/ a > = 2
=> \(\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\ge\frac{1}{4}.\left(3+2+2+2\right)=\frac{9}{4}\)
=> \(VT\le3-\frac{9}{4}=\frac{3}{4}\)
Dấu = xảy ra khi a= b = c => x = y = z
