đặt \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\), khi đó xyz = 1
bđt ⇔ \(1+\frac{3}{xy+yz+zx}\) ≥ \(\frac{6}{x+y+z}\). ta có \(\left(x+y+z\right)^2\) ≥ \(3\left(xy+yz+zx\right)\)
nên \(1+\frac{3}{xy+yz+zx}\) ≥ \(1+\frac{9}{\left(x+y+x\right)^2}\).
ta sẽ cm \(1+\frac{9}{\left(x+y+z\right)^2}\) ≥ \(\frac{6}{x+y+z}\) (*)
(*) ⇔ \(\left(1-\frac{3}{x+y+z}\right)^2\) ≥ 0 luôn đúng. do đó bđt đúng
dấu bằng xảy ra khi : x = y = z = 1 ⇔ a = b = c = 1.