When $abc=1$ find the minimum of $a^2+b^2+c^2$. When $abc=1$ find the minimum of $a^2+b^2+c^2$.
My attempt:I put $\frac{1}{bc}$ for $a$ and then I used AM-GM inequality.
$(\frac{1}{bc})^2+b^2+c^2\ge 3$
But finally we should get the answer $a+b+c$.
Where am I wrong?

AM-GM says:
$$\dfrac{a^2+b^2+c^2}3\ge\sqrt[3]{a^2b^2c^2}$$
