Let $x+y+z = 5$ where $x,y,z \in \mathbb{R}$, prove that $x^2+y^2+z^2\ge \frac{5}{3}$ My thinking:
Since $x+y+z = 5$, we can say that $x+y+z \ge  5$.
By basic fact: $x^2,y^2,z^2\ge 0$
If $x+y+z \ge  5$, then $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$
If $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$ and $x^2,y^2,z^2\ge 0$, hence $\frac{x^2+y^2+z^2}{3}\ge \frac{5}{3}$...???
I'm not sure about this, can someone help?

$(x-1)^2+(y-1)^2+(z-1)^2\ge 0$. Hence $x^2+y^2+z^2-2(x+y+z)+3\ge 0$, so $x^2+y^2+z^2\ge 10-3=7>5/3$. Moreover $(x^2+y^2+z^2)/3\ge 7/3>5/3$
Your proof is not justified: the last "hence" statement does not follow from the previous statements.
