Question

[Nguyễn Minh Chiến]

Cho x, y, z là các số thực dương. CMR:

\(\sqrt{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)\ge4\left(xy+yz+zx\right)\)

@Nguyễn Việt Lâm

Answer 1

[Võ Hồng Phúc]

Áp dụng BĐT Bunhiacopxki :

\(\left(x+y\right)\left(x+z\right)\ge\left(\sqrt{x}\sqrt{x}+\sqrt{y}\sqrt{z}\right)^2=\left(x+\sqrt{yz}\right)^2\)

\(\Rightarrow\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\)

Tương tự ta CM được:

\(\sqrt{\left(y+z\right)\left(y+x\right)}\ge y+\sqrt{xz}\) ; \(\sqrt{\left(x+z\right)\left(y+z\right)}\ge z+\sqrt{yx}\)

đặt vế trái của BĐT cần CM là A

\(\Rightarrow A=\left(x+y\right)\sqrt{\left(z+x\right)\left(z+y\right)}+\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}+\left(z+x\right)\sqrt{\left(y+z\right)\left(y+x\right)}\)

\(\ge\left(x+y\right)\left(z+\sqrt{xy}\right)+\left(y+z\right)\left(x+\sqrt{yz}\right)+\left(z+x\right)\left(y+\sqrt{zx}\right)\)

\(=\sqrt{xy}\left(x+y\right)+\sqrt{yz}\left(y+z\right)+\sqrt{zx}\left(z+x\right)+2\left(xy+yz+zx\right)\)

\(\ge2xy+2yz+2zx+2\left(xy+yz+zx\right)=4\left(xy+yz+zx\right)\)

Dấu ''='' xảy ra \(\Leftrightarrow x=y=z\)