# Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal

$x \mapsto x^2$ is a convex function. By Jensen's inequality,

$$x^2+y^2+z^2 = 3\left(\frac{x^2+y^2+z^2}{3}\right) \ge 3\left(\frac{x+y+z}{3}\right)^2 = 3\left(\frac{1}{3}\right)^2 = \frac13$$

Since the equality $x^2+y^2+z^2 = \frac13$ is achieved at $x = y = z = \frac13$, this is the solution you seek.

If you want a more elementary approach, you can use the fact

\begin{align} x^2 + y^2 + z^2 =& \left(x-\frac13+\frac13\right)^2 + \left(y-\frac13+\frac13\right)^2 + \left(z -\frac13 + \frac13\right)^2\\ =& \left(x-\frac13\right)^2 + \left(y-\frac13\right)^2 + \left(z -\frac13\right)^2\\ &+ \frac23\left[\left(x-\frac13\right) + \left(y-\frac13\right) + \left(z -\frac13\right)\right] + 3\left(\frac13\right)^2\\ =& \left(x-\frac13\right)^2 + \left(y-\frac13\right)^2 + \left(z -\frac13\right)^2 + \frac13 \end{align} to arrive at same conclusion.

By Cauchy-Schwarz, for every $x,y,z\in\Bbb R$ we have $$x+y+z=\langle(1,1,1),(x,y,z)\rangle \leq \left\|(1,1,1)\right\|_2\cdot \|(x,y,z)\|_2=\sqrt{3}\cdot \sqrt{x^2+y^2+z^2}$$ Now, if $x+y+z=1$, then squaring both sides and dividing by $3$ gives $$\frac{1}{3}\leq x^2+y^2+z^2$$ This lower bound is then attained for $|x|=|y|=|z|=1/3$.

By elementary geometry. You are looking for a point on the plane given by $x+y+z=1$ that is closest to the origin, which means that the vector $(x,y,z)$ from the origin to that point is a normal vector to the plane. This happens if $x=y=z$, which gives you three independent linear equations in $x,y,z$ to solve.