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Edit: My previous answer contained a mistake for the case where $x$ is restricted to $\mathbb{R}_{+}^{n}$. I removed this case from my answer.

Take $\bar{U} > 0$. Let's denote $\delta = (\delta_{1}, \ldots, \delta_{n})$ and $p = (p_{1}, \ldots, p_{n})$. Assume $p \neq 0$. We want to solve \begin{align*} \min_{x\in\mathbb{R}^{n}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq - \bar{U}. \end{align*}

Let's first note that a solution must satisfy $(x - \delta) \cdot (x - \delta) = \bar{U}$. This is because the left side of the constraint is continuous in $x$, and because every neighbourhood of $x$ in $\mathbb{R}^{n}$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \neq 0$).

Setting up a Lagrangian, we solve \begin{align*} \min_{x\in\mathbb{R}^{n}, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) - \bar{U} \right). \end{align*} The first order conditions are $$ p + 2 \lambda (x - \delta) = (0, \ldots, 0) $$ and $$ (x - \delta) \cdot (x - \delta) - \bar{U} = 0. $$ Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = p / (2\lambda)$$(x - \delta) = - p / (2\lambda)$. Plugging this expression for $(x - \delta)$ into the second FOC, $$ \frac{p \cdot p}{4 \lambda^{2}} = \bar{U} $$ and hence $\lambda = \sqrt{\frac{p\cdot p}{4 \bar{U}}}$. Finally, $$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{\bar{U}}{p\cdot p}} $$

To gain a geometric intuition, let's denote by $\Vert\cdot\Vert$ the Eucliden norm on $\mathbb{R}^{n}$ and observe that the set of feasible choices is $$ \left\lbrace x \in\mathbb{R}^{n}\colon \Vert\delta - x \Vert \leq \sqrt{\bar{U}}\right\rbrace. $$ This is nothing but the closed ball of radiuc $\sqrt{\bar{U}}$ around $\delta$. Minimizing the linear function $x\mapsto p \cdot x$ involves finding the point on the boundary of the ball where a level set of $x\mapsto p \cdot x$ is tangent to the ball. We find this point by traveling from the ball's center at $\delta$ in the direction of $-p$, i.e. orthogonal to the hyperplane, until we reach the boundary.

Here's an alternate proof that doesn't use any stuff involving Lagrangians: As before, we may restrict attention to points $y$ satisfying $\Vert\delta - y \Vert = \sqrt{\bar{U}}$. Any such point may be written as $y = \delta - (\delta - y)\frac{\sqrt{\bar{U}}}{\Vert \delta - y \Vert}$. Consider the point $x = \delta - p \frac{\sqrt{\bar{U}}}{\Vert p \Vert}$. We claim that $p\cdot x \leq p \cdot y$ for any such point $y$ on the boundary. This is the case if and only if \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq p\cdot p \frac{1}{\Vert p \Vert}. \end{align*} The right side equals $\Vert p \Vert$. As for the left side, by the Cauchy-Schwarz inequality, \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq \Vert p \Vert \Vert \delta - y \Vert \frac{1}{\Vert \delta - y \Vert} = \Vert p\Vert, \end{align*} and so we're done. Moreover, the Cauchy-Schwarz inequality holds strictly unless $y - \delta$ is a multiple of $p$. In that case, however, either $p = \delta - y$ or $y$ does not lie on the boundary. So the point $x$ is in fact the unique minimizer.

Edit: My previous answer contained a mistake for the case where $x$ is restricted to $\mathbb{R}_{+}^{n}$. I removed this case from my answer.

Take $\bar{U} > 0$. Let's denote $\delta = (\delta_{1}, \ldots, \delta_{n})$ and $p = (p_{1}, \ldots, p_{n})$. Assume $p \neq 0$. We want to solve \begin{align*} \min_{x\in\mathbb{R}^{n}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq - \bar{U}. \end{align*}

Let's first note that a solution must satisfy $(x - \delta) \cdot (x - \delta) = \bar{U}$. This is because the left side of the constraint is continuous in $x$, and because every neighbourhood of $x$ in $\mathbb{R}^{n}$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \neq 0$).

Setting up a Lagrangian, we solve \begin{align*} \min_{x\in\mathbb{R}^{n}, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) - \bar{U} \right). \end{align*} The first order conditions are $$ p + 2 \lambda (x - \delta) = (0, \ldots, 0) $$ and $$ (x - \delta) \cdot (x - \delta) - \bar{U} = 0. $$ Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = p / (2\lambda)$. Plugging this expression for $(x - \delta)$ into the second FOC, $$ \frac{p \cdot p}{4 \lambda^{2}} = \bar{U} $$ and hence $\lambda = \sqrt{\frac{p\cdot p}{4 \bar{U}}}$. Finally, $$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{\bar{U}}{p\cdot p}} $$

To gain a geometric intuition, let's denote by $\Vert\cdot\Vert$ the Eucliden norm on $\mathbb{R}^{n}$ and observe that the set of feasible choices is $$ \left\lbrace x \in\mathbb{R}^{n}\colon \Vert\delta - x \Vert \leq \sqrt{\bar{U}}\right\rbrace. $$ This is nothing but the closed ball of radiuc $\sqrt{\bar{U}}$ around $\delta$. Minimizing the linear function $x\mapsto p \cdot x$ involves finding the point on the boundary of the ball where a level set of $x\mapsto p \cdot x$ is tangent to the ball. We find this point by traveling from the ball's center at $\delta$ in the direction of $-p$, i.e. orthogonal to the hyperplane, until we reach the boundary.

Here's an alternate proof that doesn't use any stuff involving Lagrangians: As before, we may restrict attention to points $y$ satisfying $\Vert\delta - y \Vert = \sqrt{\bar{U}}$. Any such point may be written as $y = \delta - (\delta - y)\frac{\sqrt{\bar{U}}}{\Vert \delta - y \Vert}$. Consider the point $x = \delta - p \frac{\sqrt{\bar{U}}}{\Vert p \Vert}$. We claim that $p\cdot x \leq p \cdot y$ for any such point $y$ on the boundary. This is the case if and only if \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq p\cdot p \frac{1}{\Vert p \Vert}. \end{align*} The right side equals $\Vert p \Vert$. As for the left side, by the Cauchy-Schwarz inequality, \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq \Vert p \Vert \Vert \delta - y \Vert \frac{1}{\Vert \delta - y \Vert} = \Vert p\Vert, \end{align*} and so we're done. Moreover, the Cauchy-Schwarz inequality holds strictly unless $y - \delta$ is a multiple of $p$. In that case, however, either $p = \delta - y$ or $y$ does not lie on the boundary. So the point $x$ is in fact the unique minimizer.

Edit: My previous answer contained a mistake for the case where $x$ is restricted to $\mathbb{R}_{+}^{n}$. I removed this case from my answer.

Take $\bar{U} > 0$. Let's denote $\delta = (\delta_{1}, \ldots, \delta_{n})$ and $p = (p_{1}, \ldots, p_{n})$. Assume $p \neq 0$. We want to solve \begin{align*} \min_{x\in\mathbb{R}^{n}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq - \bar{U}. \end{align*}

Let's first note that a solution must satisfy $(x - \delta) \cdot (x - \delta) = \bar{U}$. This is because the left side of the constraint is continuous in $x$, and because every neighbourhood of $x$ in $\mathbb{R}^{n}$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \neq 0$).

Setting up a Lagrangian, we solve \begin{align*} \min_{x\in\mathbb{R}^{n}, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) - \bar{U} \right). \end{align*} The first order conditions are $$ p + 2 \lambda (x - \delta) = (0, \ldots, 0) $$ and $$ (x - \delta) \cdot (x - \delta) - \bar{U} = 0. $$ Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = - p / (2\lambda)$. Plugging this expression for $(x - \delta)$ into the second FOC, $$ \frac{p \cdot p}{4 \lambda^{2}} = \bar{U} $$ and hence $\lambda = \sqrt{\frac{p\cdot p}{4 \bar{U}}}$. Finally, $$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{\bar{U}}{p\cdot p}} $$

To gain a geometric intuition, let's denote by $\Vert\cdot\Vert$ the Eucliden norm on $\mathbb{R}^{n}$ and observe that the set of feasible choices is $$ \left\lbrace x \in\mathbb{R}^{n}\colon \Vert\delta - x \Vert \leq \sqrt{\bar{U}}\right\rbrace. $$ This is nothing but the closed ball of radiuc $\sqrt{\bar{U}}$ around $\delta$. Minimizing the linear function $x\mapsto p \cdot x$ involves finding the point on the boundary of the ball where a level set of $x\mapsto p \cdot x$ is tangent to the ball. We find this point by traveling from the ball's center at $\delta$ in the direction of $-p$, i.e. orthogonal to the hyperplane, until we reach the boundary.

Here's an alternate proof that doesn't use any stuff involving Lagrangians: As before, we may restrict attention to points $y$ satisfying $\Vert\delta - y \Vert = \sqrt{\bar{U}}$. Any such point may be written as $y = \delta - (\delta - y)\frac{\sqrt{\bar{U}}}{\Vert \delta - y \Vert}$. Consider the point $x = \delta - p \frac{\sqrt{\bar{U}}}{\Vert p \Vert}$. We claim that $p\cdot x \leq p \cdot y$ for any such point $y$ on the boundary. This is the case if and only if \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq p\cdot p \frac{1}{\Vert p \Vert}. \end{align*} The right side equals $\Vert p \Vert$. As for the left side, by the Cauchy-Schwarz inequality, \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq \Vert p \Vert \Vert \delta - y \Vert \frac{1}{\Vert \delta - y \Vert} = \Vert p\Vert, \end{align*} and so we're done. Moreover, the Cauchy-Schwarz inequality holds strictly unless $y - \delta$ is a multiple of $p$. In that case, however, either $p = \delta - y$ or $y$ does not lie on the boundary. So the point $x$ is in fact the unique minimizer.

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I wasn't sure whether or not you wantedEdit: My previous answer contained a mistake for the choice ofcase where $x$ to be constrainedis restricted to $\mathbb{R}^{n}_{+}$$\mathbb{R}_{+}^{n}$. This constraint complicates the derivation only by a little bit (see comment below)I removed this case from my answer.

Take $\bar{U} < 0$$\bar{U} > 0$. Let's denote $\delta = (\delta_{1}, \ldots, \delta_{n})$ and $p = (p_{1}, \ldots, p_{n})$. Assume $\delta > 0$ and $p \gg 0$$p \neq 0$. We want to solve \begin{align*} \min_{x\in\mathbb{R}^{n}_{+}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq \bar{U} \end{align*} Let us denote $0_{n} = (0, \ldots, 0)$ for the origin. One possibility is that $0_{n}$ is a solution (namely if $- \delta \cdot \delta \geq \bar{U}$). Suppose this is not the case (i.e. suppose $- \delta \cdot \delta < \bar{U}$). Without loss we thus consider candidates in $\mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$.\begin{align*} \min_{x\in\mathbb{R}^{n}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq - \bar{U}. \end{align*}

Let's first note that a solution must satisfy $- (x - \delta) \cdot (x - \delta) = \bar{U}$$(x - \delta) \cdot (x - \delta) = \bar{U}$. This is because the left side of the constraint is continuous in $x$, and because if $x \in \mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$, then every neighbourhood of $x$ in $\mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$$\mathbb{R}^{n}$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \gg 0$ and since $x \neq 0_{n}$$p \neq 0$).

Setting up a Lagrangian, we solve \begin{align*} \min_{x\in\mathbb{R}^{n}_{+}\setminus \lbrace 0_{n}\rbrace, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) + \bar{U} \right) \end{align*}\begin{align*} \min_{x\in\mathbb{R}^{n}, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) - \bar{U} \right). \end{align*} The first order conditions are $$ p + 2 \lambda (x - \delta) = (0, \ldots, 0) $$ and $$ (x - \delta) \cdot (x - \delta) + \bar{U} = 0. $$$$ (x - \delta) \cdot (x - \delta) - \bar{U} = 0. $$ Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = p / (2\lambda)$. Plugging this expression for $(x - \delta)$ into the second FOC, $$ \frac{p \cdot p}{4 \lambda^{2}} = - \bar{U} $$$$ \frac{p \cdot p}{4 \lambda^{2}} = \bar{U} $$ and hence $\lambda = \sqrt{\frac{p\cdot p}{- 4 \bar{U}}}$. (Recall that $-\bar{U}$ is strictly positive$\lambda = \sqrt{\frac{p\cdot p}{4 \bar{U}}}$.) Finally, $$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{-\bar{U}}{p\cdot p}} $$$$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{\bar{U}}{p\cdot p}} $$

If you want to allow for negative consumptionTo gain a geometric intuition, you can basically go through the same stepslet's denote by (and you can replace$\Vert\cdot\Vert$ the assumption $p\gg 0$ byEucliden norm on $p> 0$$\mathbb{R}^{n}$ and dropobserve that the assumptionset of feasible choices is $$ \left\lbrace x \in\mathbb{R}^{n}\colon \Vert\delta - x \Vert \leq \sqrt{\bar{U}}\right\rbrace. $$ This is nothing but the closed ball of radiuc $\delta > 0$)$\sqrt{\bar{U}}$ around $\delta$. The fact that the constraint must bind follows from Minimizing the same argument but you don't have to worry aboutlinear function $0_{n}$ being$x\mapsto p \cdot x$ involves finding the point on the boundary of the ball where a solutionlevel set of $x\mapsto p \cdot x$ is tangent to the ball. You can also allow for arbitraryWe find this point by traveling from the ball's center at $\delta$ and constrain $x$ to lie in the direction of $\mathbb{R}_{+}^{n}$$-p$, but theni.e. orthogonal to the set ofhyperplane, until we reach the boundary.

Here's an alternate proof that doesn't use any stuff involving Lagrangians: As before, we may restrict attention to points $y$ satisfying the constraint$\Vert\delta - y \Vert = \sqrt{\bar{U}}$. Any such point may be emptywritten as $y = \delta - (\delta - y)\frac{\sqrt{\bar{U}}}{\Vert \delta - y \Vert}$. Consider the point $x = \delta - p \frac{\sqrt{\bar{U}}}{\Vert p \Vert}$. We claim that $p\cdot x \leq p \cdot y$ for any such point $y$ on the boundary. This is the case if and only if \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq p\cdot p \frac{1}{\Vert p \Vert}. \end{align*} The right side equals $\bar{U}$$\Vert p \Vert$. As for the left side, by the Cauchy-Schwarz inequality, \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq \Vert p \Vert \Vert \delta - y \Vert \frac{1}{\Vert \delta - y \Vert} = \Vert p\Vert, \end{align*} and so we're done. Moreover, the Cauchy-Schwarz inequality holds strictly unless $y - \delta$ is sufficiently close to zeroa multiple of $p$. In that case, however, either $p = \delta - y$ or $y$ does not lie on the boundary. So the point $x$ is in fact the unique minimizer.

I wasn't sure whether or not you wanted the choice of $x$ to be constrained to $\mathbb{R}^{n}_{+}$. This constraint complicates the derivation only by a little bit (see comment below).

Take $\bar{U} < 0$. Let's denote $\delta = (\delta_{1}, \ldots, \delta_{n})$ and $p = (p_{1}, \ldots, p_{n})$. Assume $\delta > 0$ and $p \gg 0$. We want to solve \begin{align*} \min_{x\in\mathbb{R}^{n}_{+}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq \bar{U} \end{align*} Let us denote $0_{n} = (0, \ldots, 0)$ for the origin. One possibility is that $0_{n}$ is a solution (namely if $- \delta \cdot \delta \geq \bar{U}$). Suppose this is not the case (i.e. suppose $- \delta \cdot \delta < \bar{U}$). Without loss we thus consider candidates in $\mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$.

Let's first note that a solution must satisfy $- (x - \delta) \cdot (x - \delta) = \bar{U}$. This is because the left side of the constraint is continuous in $x$, and because if $x \in \mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$, then every neighbourhood of $x$ in $\mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \gg 0$ and since $x \neq 0_{n}$).

Setting up a Lagrangian, we solve \begin{align*} \min_{x\in\mathbb{R}^{n}_{+}\setminus \lbrace 0_{n}\rbrace, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) + \bar{U} \right) \end{align*} The first order conditions are $$ p + 2 \lambda (x - \delta) = (0, \ldots, 0) $$ and $$ (x - \delta) \cdot (x - \delta) + \bar{U} = 0. $$ Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = p / (2\lambda)$. Plugging this expression for $(x - \delta)$ into the second FOC, $$ \frac{p \cdot p}{4 \lambda^{2}} = - \bar{U} $$ and hence $\lambda = \sqrt{\frac{p\cdot p}{- 4 \bar{U}}}$. (Recall that $-\bar{U}$ is strictly positive.) Finally, $$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{-\bar{U}}{p\cdot p}} $$

If you want to allow for negative consumption, you can basically go through the same steps (and you can replace the assumption $p\gg 0$ by $p> 0$ and drop the assumption $\delta > 0$). The fact that the constraint must bind follows from the same argument but you don't have to worry about $0_{n}$ being a solution. You can also allow for arbitrary $\delta$ and constrain $x$ to lie in $\mathbb{R}_{+}^{n}$, but then the set of points satisfying the constraint may be empty if $\bar{U}$ is sufficiently close to zero.

Edit: My previous answer contained a mistake for the case where $x$ is restricted to $\mathbb{R}_{+}^{n}$. I removed this case from my answer.

Take $\bar{U} > 0$. Let's denote $\delta = (\delta_{1}, \ldots, \delta_{n})$ and $p = (p_{1}, \ldots, p_{n})$. Assume $p \neq 0$. We want to solve \begin{align*} \min_{x\in\mathbb{R}^{n}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq - \bar{U}. \end{align*}

Let's first note that a solution must satisfy $(x - \delta) \cdot (x - \delta) = \bar{U}$. This is because the left side of the constraint is continuous in $x$, and because every neighbourhood of $x$ in $\mathbb{R}^{n}$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \neq 0$).

Setting up a Lagrangian, we solve \begin{align*} \min_{x\in\mathbb{R}^{n}, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) - \bar{U} \right). \end{align*} The first order conditions are $$ p + 2 \lambda (x - \delta) = (0, \ldots, 0) $$ and $$ (x - \delta) \cdot (x - \delta) - \bar{U} = 0. $$ Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = p / (2\lambda)$. Plugging this expression for $(x - \delta)$ into the second FOC, $$ \frac{p \cdot p}{4 \lambda^{2}} = \bar{U} $$ and hence $\lambda = \sqrt{\frac{p\cdot p}{4 \bar{U}}}$. Finally, $$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{\bar{U}}{p\cdot p}} $$

To gain a geometric intuition, let's denote by $\Vert\cdot\Vert$ the Eucliden norm on $\mathbb{R}^{n}$ and observe that the set of feasible choices is $$ \left\lbrace x \in\mathbb{R}^{n}\colon \Vert\delta - x \Vert \leq \sqrt{\bar{U}}\right\rbrace. $$ This is nothing but the closed ball of radiuc $\sqrt{\bar{U}}$ around $\delta$. Minimizing the linear function $x\mapsto p \cdot x$ involves finding the point on the boundary of the ball where a level set of $x\mapsto p \cdot x$ is tangent to the ball. We find this point by traveling from the ball's center at $\delta$ in the direction of $-p$, i.e. orthogonal to the hyperplane, until we reach the boundary.

Here's an alternate proof that doesn't use any stuff involving Lagrangians: As before, we may restrict attention to points $y$ satisfying $\Vert\delta - y \Vert = \sqrt{\bar{U}}$. Any such point may be written as $y = \delta - (\delta - y)\frac{\sqrt{\bar{U}}}{\Vert \delta - y \Vert}$. Consider the point $x = \delta - p \frac{\sqrt{\bar{U}}}{\Vert p \Vert}$. We claim that $p\cdot x \leq p \cdot y$ for any such point $y$ on the boundary. This is the case if and only if \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq p\cdot p \frac{1}{\Vert p \Vert}. \end{align*} The right side equals $\Vert p \Vert$. As for the left side, by the Cauchy-Schwarz inequality, \begin{align*} p\cdot (\delta - y) \frac{1}{\Vert \delta - y \Vert} \leq \Vert p \Vert \Vert \delta - y \Vert \frac{1}{\Vert \delta - y \Vert} = \Vert p\Vert, \end{align*} and so we're done. Moreover, the Cauchy-Schwarz inequality holds strictly unless $y - \delta$ is a multiple of $p$. In that case, however, either $p = \delta - y$ or $y$ does not lie on the boundary. So the point $x$ is in fact the unique minimizer.

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induction_is_a_laddah
induction_is_a_laddah
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I wasn't sure whether or not you wanted the choice of $x$ to be constrained to $\mathbb{R}^{n}_{+}$. This constraint complicates the derivation only by a little bit (see comment below).

Take $\bar{U} < 0$. Let's denote $\delta = (\delta_{1}, \ldots, \delta_{n})$ and $p = (p_{1}, \ldots, p_{n})$. Assume $\delta > 0$ and $p \gg 0$. We want to solve \begin{align*} \min_{x\in\mathbb{R}^{n}_{+}} p \cdot x \qquad \text{s.t.}\quad - (x - \delta) \cdot (x - \delta) \geq \bar{U} \end{align*} Let us denote $0_{n} = (0, \ldots, 0)$ for the origin. One possibility is that $0_{n}$ is a solution (namely if $- \delta \cdot \delta \geq \bar{U}$). Suppose this is not the case (i.e. suppose $- \delta \cdot \delta < \bar{U}$). Without loss we thus consider candidates in $\mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$.

Let's first note that a solution must satisfy $- (x - \delta) \cdot (x - \delta) = \bar{U}$. This is because the left side of the constraint is continuous in $x$, and because if $x \in \mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$, then every neighbourhood of $x$ in $\mathbb{R}_{+}^{n}\setminus \lbrace 0_{n}\rbrace$ contains a point $x^{\prime}$ such that $p\cdot x^{\prime} < p \cdot x$ (since $p \gg 0$ and since $x \neq 0_{n}$).

Setting up a Lagrangian, we solve \begin{align*} \min_{x\in\mathbb{R}^{n}_{+}\setminus \lbrace 0_{n}\rbrace, \lambda\in\mathbb{R}} p \cdot x + \lambda\left( (x - \delta) \cdot (x - \delta) + \bar{U} \right) \end{align*} The first order conditions are $$ p + 2 \lambda (x - \delta) = (0, \ldots, 0) $$ and $$ (x - \delta) \cdot (x - \delta) + \bar{U} = 0. $$ Note that $\lambda$ cannot equal zero as otherwise the first FOC is contradicted. The first FOC is therefore equivalent to $(x - \delta) = p / (2\lambda)$. Plugging this expression for $(x - \delta)$ into the second FOC, $$ \frac{p \cdot p}{4 \lambda^{2}} = - \bar{U} $$ and hence $\lambda = \sqrt{\frac{p\cdot p}{- 4 \bar{U}}}$. (Recall that $-\bar{U}$ is strictly positive.) Finally, $$ x = \delta + p \frac{1}{2\delta} = \delta + p \sqrt{\frac{-\bar{U}}{p\cdot p}} $$

If you want to allow for negative consumption, you can basically go through the same steps (and you can replace the assumption $p\gg 0$ by $p> 0$ and drop the assumption $\delta > 0$). The fact that the constraint must bind follows from the same argument but you don't have to worry about $0_{n}$ being a solution. You can also allow for arbitrary $\delta$ and constrain $x$ to lie in $\mathbb{R}_{+}^{n}$, but then the set of points satisfying the constraint may be empty if $\bar{U}$ is sufficiently close to zero.