2021/088) For real x,y z if x +y + z = 3 prove that $x^2+y^2 + z^2 >= 3$

We are given 

$x+y+z = 3 \cdots(1)$

now $(x-1)^2\ge 0$

or $x^2-2x + 1 \ge 0$

or $x^2 \ge 2x -1\cdots(2)$

similarly $y^2 \ge 2y-1\cdots(3)$

and $z^2\ge 2z -1\cdots(4)$

addding (2), (3), (4) we get $x^2+y^2 + z^2\ge 2(x+y+z) - 3$

or $x^2+y^2 + z^2\ge 2* 3- 3$ (from (1)

or $x^2+y^2+z^2 \ge 3$