Question

Chứng minh rằng\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)

Answer 1

Chứng minh:

Ta có:

\(\left(x-y\right)^2\ge0\Rightarrow x^2+y^2-2xy\ge0\Rightarrow x^2+y^2\ge2xy\)

\(\left(y-z\right)^2\ge0\Rightarrow y^2+z^2-2yz\ge0\Rightarrow y^2+z^2\ge2yz\)

\(\left(x-z\right)^2\ge0\Rightarrow x^2+z^2-2xz\ge0\Rightarrow x^2+z^2\ge2xz\)

Cộng vế với vế, ta được:

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

\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

\(\Rightarrow x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)(đpcm)

Answer 2

\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)

\(\Leftrightarrow x^2+y^2+z^2-\frac{1}{3}\cdot\left(x+y+z\right)^2\ge0\)

\(\Leftrightarrow x^2+y^2+z^2-\frac{1}{3}\left(x^2+y^2+z^2+2xy+2yz+2xz\right)\ge0\)

\(\Leftrightarrow x^2+y^2+z^2-\frac{1}{3}\left(x^2+y^2+z^2\right)-\frac{2}{3}\left(xy+yz+zx\right)\ge0\)

\(\Leftrightarrow\frac{2}{3}\left(x^2+y^2+z^2\right)-\frac{2}{3}\left(xy+yz+xz\right)\ge0\)

\(\Leftrightarrow\frac{2}{3}\left(x^2+y^2+z^2-xy-yz-xz\right)\ge0\) (1)

Ta cần chứng minh : \(x^2+y^2+z^2-xy-yz-xz\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)

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

=> bđt (1) đúng

\(\Rightarrow x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\) (đpcm)