Let $A \in\mathbb{R}^{m\times n}$, $b,c\in\mathbb{R}^m$, $x\in \mathbb{R}^n$. Consider the following minimization problem: $$ \min_{x\succeq 0} f(x):= \frac{1}{2}\|Ax-c\|^2 + b^\top Ax. $$ For the unconstrained case $x\in\mathbb{R}^n$, Since the function is quadratic, and in case $A^\top A$ is invertible, the solution is found by imposing $$ \nabla_x f(x) = 0 \iff x^\star = (A^\top A)^{-1}(A^\top c - A^\top b). $$ If we add the constraint $x\succeq 0$, is $x^\star_{+} = \max\{0,x^\star\}$ a solution? ($\max$ is the element-wise maximum operator here).
Related question: Simple Quadratic Minimization Problem
The answer is no. As an example, take $$ A = \pmatrix{2&1\\1&2}, \quad b = 0, \quad c = \pmatrix{1\\-1}. $$ The solution to the unconstrained problem is $x^\star = (1,-1)$, but the constrained problem has solution $x = (1/5,0)$.