Let $V > 0$ and let $\Phi(\cdot)$ be the standard normal CDF.


Consider the infimum of
$$f(x\_1, x\_2,x\_3, p\_1, p\_2, p\_3) := p\_1 \Phi(x\_1) + p\_2 \Phi(x\_2) + p\_3 \Phi(x\_3)$$
with respect to $x\_1, x\_2, x\_3, p\_1, p\_2$ and $p\_3$ satisfying


$\begin{align}
&p\_1 \geq 0,\\\\
&p\_2 \geq 0,\\\\
&p\_3 \geq 0,\\\\
&p\_1 + p\_2 + p\_3 = 1,\\\\
&p\_1 x\_1 + p\_2 x\_2 + p\_3 x\_3 = 0,\\\\
&p\_1 x\_1^2 + p\_2 x\_2^2 + p\_3 x\_3^2 \leq V. 
\end{align}
$


The objective is a continuous function and the feasible set is a subset of $\mathbb{R}^6$. One way to show that the infimum is a minimum is to show that the feasible set is compact. But even with the constraint $p\_1 x\_1^2 + p\_2 x\_2^2 + p\_3 x\_3^2 \leq V$, it can happen that $p\_1$ is arbitrarily small and $x\_1$ is arbitrarily large.


I think I need to use the fact that the objective function is bounded between $0$ and $1$, which together with the last constraint limits $x\_i$'s becoming very large.



Showing that the infimum is a minimum
$\newcommand{\R}{\mathbb R}\newcommand{\tx}{\tilde x}$Indeed, the infimum (say $f^\*$) in question is attained for each $V\in(0,\infty)$.


To prove this, for each $j\in\{1,2,3\}$ let $(x\_j^n)\_{n=1}^\infty$ and $(p\_j^n)\_{n=1}^\infty$ be sequences such that for each $n$ all your conditions are satisfied with $x\_1^n,x\_2^n,x\_3^n,p\_1^n,p\_2^n,p\_3^n$ in place of $x\_1,x\_2,x\_3,p\_1,p\_2,p\_3$ and
\begin{equation}
 f(x\_1^n,x\_2^n,x\_3^n,p\_1^n,p\_2^n,p\_3^n)\to f^\*
\end{equation}
(as $n\to\infty$). Passing to subsequences, without loss of generality assume that for each $j\in\{1,2,3\}$
\begin{equation}
x\_j^n\to x\_j^\*\in[-\infty,\infty],\quad p\_j^n\to p\_j^\*\in[0,1]. 
\end{equation}
Let
\begin{equation}
 J:=\{j\in\{1,2,3\}\colon x\_j^\*\in\R\},\quad J^c:=\{1,2,3\}\setminus J. 
\end{equation}
Since $p\_j^n (x\_j^n)^2\le V<\infty$, we see that for each $j\in J^c$ we have $p\_j^\*=0$ and $p\_j^n x\_j^n\to0$. So, letting $\tx\_j:=x\_j^\*\,1(j\in J)$ for $j\in\{1,2,3\}$, we have
\begin{equation}
 p\_j^\*\ge0\ \forall j\in\{1,2,3\},\quad \sum\_{j=1}^3 p\_j^\*=1, 
\end{equation}
\begin{equation}
 \sum\_{j=1}^3 p\_j^\*\tx\_j=\sum\_{j\in J} p\_j^\*x\_j^\*=\lim\_n \sum\_{j=1}^3 p\_j^n x\_j^n=\lim\_n 0=0, 
\end{equation}
\begin{equation}
 \sum\_{j=1}^3 p\_j^\*\tx\_j^2=\sum\_{j\in J} p\_j^\*(x\_j^\*)^2=\lim\_n \sum\_{j\in J}p\_j^n (x\_j^n)^2
 \le\limsup\_n \sum\_{j=1}^3 p\_j^n (x\_j^n)^2\le V, 
\end{equation}
\begin{equation}
 \sum\_{j=1}^3 p\_j^\*\Phi(\tx\_j)=\sum\_{j\in J} p\_j^\*\Phi(x\_j^\*)
 =\lim\_n \sum\_{j\in J} p\_j^n \Phi(x\_j^n)
 =\lim\_n \sum\_{j=1}^3 p\_j^n \Phi(x\_j^n)=f^\*. 
\end{equation}


So, all your conditions are satisfied with $\tx\_1,\tx\_2,\tx\_3,p\_1^\*,p\_2^\*,p\_3^\*$ in place of $x\_1,x\_2,x\_3,p\_1,p\_2,p\_3$ and
the infimum $f^\*$ is attained at $(x\_1,x\_2,x\_3,p\_1,p\_2,p\_3)=(\tx\_1,\tx\_2,\tx\_3,p\_1^\*,p\_2^\*,p\_3^\*)$.
$\quad\Box$


