= Solution
A useful form of the <Lagrangian sufficiency theorem> is as follows. For minimization of $f(x)$ subject to $g_j(x)\geq0$ and $h_i(x)=0$, form
$$
L(x,\lambda,\nu)=f(x)-\sum_j\lambda_jg_j(x)+\sum_i\nu_i h_i(x).
$$
If $x_*$ is feasible, $\lambda_j\geq0$, $\lambda_jg_j(x_*)=0$, and $x_*$ globally minimizes $L$ over the ambient domain for these fixed multipliers, then $x_*$ is a global constrained minimizer. Indeed, for every feasible $x$,
$$
f(x)\geq L(x,\lambda,\nu)\geq L(x_*,\lambda,\nu)=f(x_*).
$$
The first inequality uses multiplier signs; the last equality uses feasibility and <complementary slackness>. Equality constraints alone require no sign restriction on their multipliers. A stationary point of a differentiable <convex> Lagrangian is a global minimum, which is a common way to verify the theorem's hypothesis.
Here take
$$
L=2x_1^2+2x_2^2+x_3^2-\lambda(x_1+x_2+x_3-1).
$$
Stationarity gives $4x_1=4x_2=2x_3=\lambda$. Feasibility then gives $\lambda=1$ and $x_*=(1/4,1/4,1/2)$. For that multiplier,
$$
L=2(x_1-1/4)^2+2(x_2-1/4)^2+(x_3-1/2)^2+\tfrac12.
$$
The Lagrangian has its unique global minimum at $x_*$, so the sufficiency theorem proves
$$
\boxed{x_*=(1/4,1/4,1/2),\qquad \min f=1/2.}
$$
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