Proving the non-existence of a sequence satisfying a set of inequalities There exist some type of conditions called sequential optimality conditions that usually require some inequalities to be ensured along with it,  see short introdution. Based on a conjecture on this type of sequences, It would be enlightening to this conjecture to find an analytic function $\boldsymbol{c} : \mathbb{R}^{n} \rightarrow \mathbb{R}^{p}$ and sequences $\{\boldsymbol{\mu}^{k}\} \subset \mathbb{R}^p_{+}$, which means $\mu^k_i \geq 0$ for all $i\leq p$ and $k\in\mathbb{N}$, and a convergent sequence $\{\boldsymbol{x}^{k}\} \subset \mathbb{R}^n$ such that, for all $k\in\mathbb{N}$, it's true that $\boldsymbol{c}(\boldsymbol{x}^{k}) \geq \boldsymbol{0}$,
\begin{gather*}
\left( \sum_{i=1}^{p} c_{i}(\boldsymbol{x}^{k}) \mu_i^{k}\right)^{2\theta} \left( \sum_{i=1}^{p}\mu_i^{k}\right)^{2(1-\theta)} \leq M + \left( \sum_{i=1}^{p} (c_{i}(\boldsymbol{x}^{k}))^2 \right) \left( \sum_{i=1}^{p}(\mu_i^{k})^2\right),\\
\left( \sum_{i=1}^{p} c_{i}(\boldsymbol{x}^{k}) \mu_i^{k}\right) \geq \delta
\end{gather*} and
$$
\lim_{k\rightarrow \infty} \sum_{i=1}^{p} c_{i}(\boldsymbol{x}^{k}) = 0
$$ for some $\delta>0$ and $M>0$ and all $0<\theta<1$. Is it possible?
 A: $\def\paren#1{\left(#1\right)}$The answer is negative. Suppose it is possible, define\begin{gather*}
y_{k, n} = c_k(x_{1, n}, \cdots, x_{p, n}),\quad Y_n = \max_{1 \leqslant k \leqslant p} y_{k, n},\quad M_n = \max_{1 \leqslant k \leqslant p} μ_{k, n},\quad \forall 1 \leqslant k \leqslant p,\ n \geqslant 1
\end{gather*}
then the conditions are $μ_{k, n}, y_{k, n} \geqslant 0$ and\begin{gather*}
\small\paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} }^{2θ} \paren{ \sum_{k = 1}^p μ_{k, n} }^{2(1 - θ)} \leqslant M + \paren{ \sum_{k = 1}^p y_{k, n}^2 } \paren{ \sum_{k = 1}^p μ_{k, n}^2 },\quad \forall 0 < θ < 1,\ n \geqslant 1 \tag{1}\\
\sum_{k = 1}^p y_{k, n} μ_{k, n} \geqslant δ,\quad \forall n\geqslant 1\tag{2}\\
\lim_{n → ∞} \sum_{k = 1}^p y_{k, n} = 0. \tag{3}
\end{gather*}
First, because\begin{align*}
&\mathrel{\phantom{=}} \sup_{θ \in (0, 1)} \ln\paren{ \paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} }^{2θ} \paren{ \sum_{k = 1}^p μ_{k, n} }^{2(1 - θ)} }\\
&= 2\sup_{θ \in (0, 1)} \paren{ θ\ln\paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} } + (1 - θ)\ln\paren{ \sum_{k = 1}^p μ_{k, n} } }\\
&= 2\max\paren{ \ln\paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} }, \ln\paren{ \sum_{k = 1}^p μ_{k, n} } },
\end{align*}
and by the Cauchy-Schwarz inequality,$$
\paren{ \sum_{k = 1}^p y_{k, n} μ_{k, n} }^2 \leqslant \paren{ \sum_{k = 1}^p y_{k, n}^2 } \paren{ \sum_{k = 1}^p μ_{k, n}^2 },
$$
so (1) is equivalent to\begin{gather*}
\paren{ \sum_{k = 1}^p μ_{k, n} }^2 \leqslant M + \paren{ \sum_{k = 1}^p y_{k, n}^2 } \paren{ \sum_{k = 1}^p μ_{k, n}^2 }. \quad \forall n \geqslant 1 \tag{1$'$}
\end{gather*}
Next, since $\sum\limits_{k = 1}^p y_{k, n} μ_{k, n} \leqslant pY_n M_n$ and $0 \leqslant Y_n \leqslant \sum\limits_{k = 1}^p y_{k, n}$, then$$
M_n \geqslant \frac{δ}{pY_n},\quad \lim_{n → ∞} Y_n = 0,
$$
from (2) and (3), respectively. However by (1'),$$
M_n^2 \leqslant \paren{ \sum_{k = 1}^p μ_{k, n} }^2 \leqslant M + \paren{ \sum_{k = 1}^p y_{k, n}^2 } \paren{ \sum_{k = 1}^p μ_{k, n}^2 } \leqslant M + p^2 Y_n^2 M_n^2,
$$
thus for sufficiently large $n$ such that $Y_n < \dfrac{1}{p}$,$$
M \geqslant (1 - p^2 Y_n^2) M_n^2 \geqslant (1 - p^2 Y_n^2) \frac{δ^2}{p^2 Y_n^2} = δ^2 \paren{ \frac{1}{p^2 Y_n^2} - 1 },
$$
which is contradictory with $\lim\limits_{n → ∞} Y_n = 0$.
