[proofplan] The reverse implication is immediate from substituting points satisfying $q_1(x) \le 0$ into the multiplier inequality. For the converse, we homogenize both quadratic functions by writing them as quadratic forms in $z = (z_0,z') \in \mathbb{R}^{n+1}$, then use Dines' convexity theorem for the joint range of two homogeneous quadratic forms. The hypothesis $q_1(\bar{x}) < 0$ rules out a degenerate separating hyperplane and forces the separator to have the form $(\lambda,1)$ with $\lambda \ge 0$. Evaluating the resulting homogeneous inequality at $z = (1,x)$ gives the desired global nonnegativity of $q_0+\lambda q_1$. [/proofplan]

[step:Derive the implication from a nonnegative multiplier] Assume that there exists $\lambda \in [0,\infty)$ such that $q_0(x)+\lambda q_1(x) \ge 0$ for every $x \in \mathbb{R}^n$. Let $x \in \mathbb{R}^n$ satisfy $q_1(x) \le 0$. Since $\lambda \ge 0$, we have $\lambda q_1(x) \le 0$, and therefore \begin{align*} q_0(x) \ge q_0(x)+\lambda q_1(x) \ge 0. \end{align*} Thus $q_1(x) \le 0$ implies $q_0(x) \ge 0$ for every $x \in \mathbb{R}^n$. [/step]

[step:Homogenize the two quadratic functions]Assume conversely that \begin{align*} q_1(x) \le 0 \implies q_0(x) \ge 0 \end{align*} for every $x \in \mathbb{R}^n$. Since $q_i$ is a real quadratic polynomial function, there are a symmetric matrix $A_i \in \mathbb{R}^{n \times n}$, a vector $b_i \in \mathbb{R}^n$, and a scalar $c_i \in \mathbb{R}$ such that \begin{align*} q_i(x)=x^\top A_i x+2b_i^\top x+c_i \end{align*} for every $x \in \mathbb{R}^n$, for $i \in \{0,1\}$. Define the symmetric matrix $Q_i \in \mathbb{R}^{(n+1)\times(n+1)}$ by specifying its entries as follows: $(Q_i)_{00}=c_i$, $(Q_i)_{0j}=(b_i)_j$, $(Q_i)_{j0}=(b_i)_j$, and $(Q_i)_{jk}=(A_i)_{jk}$ for all $j,k\in\{1,\dots,n\}$. For $z=(z_0,z') \in \mathbb{R}\times\mathbb{R}^n$, define the homogeneous quadratic forms $p_i:\mathbb{R}^{n+1}\to\mathbb{R}$ by \begin{align*} p_i(z)=z^\top Q_i z. \end{align*} Then, for every $x \in \mathbb{R}^n$ and every $i \in \{0,1\}$, \begin{align*} p_i(1,x)=q_i(x). \end{align*} More generally, if $z=(z_0,z')$ with $z_0 \ne 0$, then \begin{align*} p_i(z)=z_0^2 q_i(z'/z_0). \end{align*}[/step]

[guided]The purpose of homogenization is to replace affine quadratic functions on $\mathbb{R}^n$ by genuine homogeneous quadratic forms on one higher-dimensional space. Write each quadratic polynomial $q_i:\mathbb{R}^n\to\mathbb{R}$ in the form \begin{align*} q_i(x)=x^\top A_i x+2b_i^\top x+c_i, \end{align*} where $A_i \in \mathbb{R}^{n\times n}$ is symmetric, $b_i \in \mathbb{R}^n$, and $c_i \in \mathbb{R}$. The symmetry of $A_i$ is harmless because the skew-symmetric part of a matrix contributes nothing to $x^\top A_i x$. Now define the symmetric matrix $Q_i\in\mathbb{R}^{(n+1)\times(n+1)}$ by $(Q_i)_{00}=c_i$, $(Q_i)_{0j}=(b_i)_j$, $(Q_i)_{j0}=(b_i)_j$, and $(Q_i)_{jk}=(A_i)_{jk}$ for all $j,k\in\{1,\dots,n\}$. Then define $p_i:\mathbb{R}^{n+1}\to\mathbb{R}$ by \begin{align*} p_i(z)=z^\top Q_i z. \end{align*} For $z=(1,x)$, this gives \begin{align*} p_i(1,x)=c_i+2b_i^\top x+x^\top A_i x=q_i(x). \end{align*} For a general vector $z=(z_0,z')$ with $z_0\ne 0$, factor out $z_0^2$: \begin{align*} p_i(z)=z_0^2\left((z'/z_0)^\top A_i(z'/z_0)+2b_i^\top(z'/z_0)+c_i\right). \end{align*} Hence \begin{align*} p_i(z)=z_0^2q_i(z'/z_0). \end{align*} This identity is the bridge between the original implication on $\mathbb{R}^n$ and the geometry of homogeneous quadratic forms on $\mathbb{R}^{n+1}$.[/guided]

[step:Use the original implication to exclude the forbidden quadrant in the homogenized joint range] Define the joint range cone $C \subset \mathbb{R}^2$ by \begin{align*} C=\{(p_1(z),p_0(z)) : z \in \mathbb{R}^{n+1}\}. \end{align*} By Dines' convexity theorem for two real quadratic forms, $C$ is a convex cone in $\mathbb{R}^2$ (citing a result not yet in the wiki: Dines' convexity theorem for the joint range of two quadratic forms). We claim that \begin{align*} C\cap\{(a,b)\in\mathbb{R}^2:a\le 0,\ b<0\}=\varnothing. \end{align*} First take $z=(z_0,z')$ with $z_0\ne 0$ and set $x=z'/z_0$. If $p_1(z)\le 0$, then $z_0^2q_1(x)\le 0$, hence $q_1(x)\le 0$. By the hypothesis, $q_0(x)\ge 0$, and so \begin{align*} p_0(z)=z_0^2q_0(x)\ge 0. \end{align*} Thus no vector with $z_0\ne 0$ gives a point $(p_1(z),p_0(z))$ with first coordinate nonpositive and second coordinate negative. It remains to exclude vectors with $z_0=0$. Suppose, to the contrary, that $z\in\mathbb{R}^{n+1}$ satisfies \begin{align*} p_1(z)\le 0 \end{align*} and \begin{align*} p_0(z)<0. \end{align*} Choose the strict feasible point $\bar{x}\in\mathbb{R}^n$ from the hypothesis and define $\bar{z}=(1,\bar{x})\in\mathbb{R}^{n+1}$. Then \begin{align*} p_1(\bar{z})=q_1(\bar{x})<0. \end{align*} Since $p_0(z)<0$, there exists $\varepsilon>0$ small enough that \begin{align*} p_0(z)+\varepsilon p_0(\bar{z})<0. \end{align*} Because $C$ is a convex cone, the sum of two elements of $C$ belongs to $C$, so \begin{align*} (p_1(z),p_0(z))+\varepsilon(p_1(\bar{z}),p_0(\bar{z}))\in C. \end{align*} Its first coordinate is strictly negative and its second coordinate is strictly negative. Therefore there exists $w\in\mathbb{R}^{n+1}$ such that \begin{align*} p_1(w)<0 \end{align*} and \begin{align*} p_0(w)<0. \end{align*} If $w_0\ne 0$, this contradicts the previous paragraph. If $w_0=0$, choose a sequence $(w_k)_{k\in\mathbb{N}}$ in $\mathbb{R}^{n+1}$ with nonzero first coordinate such that $w_k\to w$ in $\mathbb{R}^{n+1}$. Since $p_0$ and $p_1$ are continuous, for all sufficiently large $k$ we have $p_1(w_k)<0$ and $p_0(w_k)<0$, again contradicting the previous paragraph. Hence the claimed exclusion holds. [/step]

[step:Separate the joint range from the forbidden quadrant] Let \begin{align*} K=\{(a,b)\in\mathbb{R}^2:a\le 0,\ b<0\}. \end{align*} The previous step gives $C\cap K=\varnothing$. Since $C$ is a convex cone and $K$ is a convex cone with nonempty interior, the finite-dimensional separating hyperplane theorem gives a nonzero vector $(\alpha,\beta)\in\mathbb{R}^2$ such that \begin{align*} \alpha a+\beta b\ge 0 \end{align*} for every $(a,b)\in C$, while \begin{align*} \alpha a+\beta b\le 0 \end{align*} for every $(a,b)\in K$ (citing a result not yet in the wiki: finite-dimensional separating hyperplane theorem for disjoint convex cones). The inequality on $K$ forces $\alpha\ge 0$ and $\beta\ge 0$: if $\alpha<0$, then taking $a<0$ with large magnitude and a fixed $b<0$ makes $\alpha a+\beta b>0$; if $\beta<0$, then taking $b<0$ with large magnitude and a fixed $a\le 0$ makes $\alpha a+\beta b>0$. Moreover $\beta>0$. Indeed, if $\beta=0$, then $\alpha>0$ and the inequality on $C$ gives $\alpha p_1(z)\ge 0$ for every $z\in\mathbb{R}^{n+1}$. Applying this to $\bar{z}=(1,\bar{x})$ gives \begin{align*} \alpha q_1(\bar{x})=\alpha p_1(\bar{z})\ge 0, \end{align*} contradicting $\alpha>0$ and $q_1(\bar{x})<0$. Therefore $\beta>0$. Define \begin{align*} \lambda=\alpha/\beta. \end{align*} Then $\lambda\ge 0$, and for every $z\in\mathbb{R}^{n+1}$ we have \begin{align*} \lambda p_1(z)+p_0(z)\ge 0. \end{align*} [/step]

[step:Evaluate the separated homogeneous inequality on the affine slice] For every $x\in\mathbb{R}^n$, apply the inequality from the previous step to $z=(1,x)$. Since $p_i(1,x)=q_i(x)$ for $i\in\{0,1\}$, we obtain \begin{align*} q_0(x)+\lambda q_1(x)=p_0(1,x)+\lambda p_1(1,x)\ge 0. \end{align*} The scalar $\lambda$ belongs to $[0,\infty)$, so this is precisely the required multiplier certificate. Combining this converse implication with the first step proves the equivalence. [/step]