Hey everyone,
I need some help on finding the min value of x^2+y^2+z^2 if x + y + z is 1
Thanks.
$\displaystyle \begin{array}{rcl} \left(x-\dfrac13\right)^2+\left(y-\dfrac13\right)^2+\left(z-\dfrac13\right)^2 & \geqslant & 0 \\\\ x^2+y^2+z^2 & \geqslant & \dfrac23(x+y+z)-3\left(\dfrac19\right) \\\\ {} & = & \dfrac13 \end{array}$
The formula given above sees correct for the problem asked. this has calculated the exact amount of min value.
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