Question

[Nguyễn Văn Quang]

∀ x, y, z, cmr:

a) x² + y² + z² ≥ xy + yz + zx

b) x² + y² + z² + 3 ≥ 2(x + y +z)

c) \(\frac{x^2+y^2+z^2}{3}\ge\left(\frac{x+y+z}{3}\right)^3\)

Answer 1

[Nguyễn Việt Lâm]

a/ \(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

\(\Leftrightarrow x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(x=y=z\)

b/ \(\Leftrightarrow x^2-2x+1+y^2-2y+1+z^2-2z+1\ge0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(x=y=z=1\)

c/ BĐT sai