<s>[INST] 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?
 [/INST] Viewed geometrically, $x+y+z=5$ is a plane and $x^2+y^2+z^2=a^2$ is a sphere with radius $a$ at the origin; the smallest (positive) $a$ for which the sphere touches the plane corresponds to the shortest distance between the plane and the origin, and the point of tangency represents the smallest possible value of $x^2+y^2+z^2$. Intuitively this is achieved at $x=y=z=\frac53$, whence the result follows (indeed $x^2+y^2+z^2\ge\frac{25}3$).
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