Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality Using the AM and GM inequality, given that
$a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2\geqslant\frac{1}{3}$$

$$a^2+{1\over 9} + b^2+{1\over 9} + c^2+{1\over 9}\geq {2\over 3}(a+b+c)$$ by AM-GM.
