Question

[Họ Và Tên]

cho x,y,z>0

chứng minh rằng 

\(\sqrt{x^2+xy+2y^2}+\sqrt{y^2+yz+2z^2}+\sqrt{z^2+zx+2x^2}\ge2\left(x+y+z\right)\)

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow\sqrt{4x^2+4xy+8y^2}+\sqrt{4y^2+4yz+8z^2}+\sqrt{4z^2+4zx+8x^2}\ge4\left(x+y+z\right)\)

Ta có:

\(VT=\sqrt{\left(2x+y\right)^2+\left(\sqrt{7}y\right)^2}+\sqrt{\left(2y+z\right)^2+\left(\sqrt{7}z\right)^2}+\sqrt{\left(2z+x\right)^2+\left(\sqrt{7}x\right)^2}\)

\(VT\ge\sqrt{\left(2x+y+2y+z+2z+x\right)^2+\left(\sqrt{7}x+\sqrt{7}y+\sqrt{7}z\right)^2}\)

\(VT\ge\sqrt{16\left(x+y+z\right)^2}=4\left(x+y+z\right)\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\)

Answer 2

[Nguyễn Việt Lâm]

BĐT Mincopxki:

\(\sqrt{x^2+a^2}+\sqrt{y^2+b^2}+\sqrt{z^2+c^2}\ge\sqrt{\left(x+y+z\right)^2+\left(a+b+c\right)^2}\)