Let m be the smallest number among: $(x-y)^2, (y-z)^2, (z-x)^2$ Prove $m \le \frac{1}{2}(x^2+y^2+z^2)$ Let m be the smallest number among: $(x-y)^2, (y-z)^2, (z-x)^2$
Prove $m \le \frac{1}{2}(x^2+y^2+z^2)$
My attempts: $3m \le (x-y)^2+(y-z)^2+(z-x)^2$ so I tried to prove $(x-y)^2+(y-z)^2+(z-x)^2 \le \frac{3}{2}(x^2+y^2+z^2) \rightarrow 4(xy+yz+zx)\le x^2+y^2+z^2$ but it wrong
Can anyone help me with this problem? Thank you

Wlog assume that $x\ge y \ge z $ , then  $m=min\{(x-y)^2, (y-z)^2\} \le (x-y)(y-z)$
Note that
$$x^2+y^2+z^2-2(x-y)(y-z)=(x-y+z)^2+2y^2 \ge 0$$
Hence
$$x^2+y^2+z^2 \ge 2(x-y)(y-z) \ge 2m$$
Equality holds when $y=0, x=-z$
