<s>[INST] 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
$$a^2+b^2+c^2\geqslant\frac{1}{3}$$
 [/INST] In the worst case possible you'd get $$a = b = c = \frac{1}{3} \Longrightarrow a^2 + b^2 + c^2 = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} \geq \frac{1}{3} $$ 
In the best case possible you'd get $$a = 1, b = c = 0 \Longrightarrow 1^2 + 0^2 + 0^2 = 1 \geq 1/3 $$
Therefore the inequality holds. Didn't use the AM-GM inequality, though.
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