Consider the primal problem (P)
(P) $f^*$= $\min f(x)$
s.t $g_i(x)\le0$
where $f,g_i$ are convex functoin and suppose there exists $\hat{x}$ such that $g_i(\hat{x})<0$ and (P) is bounded from below ($f^*>-\infty$).
Consider the dual problem given by $\max\{q(\lambda):\lambda\in dom(q)\}$
where $q(\lambda)=\min_xL(x,\lambda),dom(q)=\{\lambda\in\mathbb{R}^m_+:q(\lambda)>-\infty\}$,Let $\lambda^*$be an optimal solution of the dual problem.
Prove
$$\sum_{i=1}^{m}\lambda_i^*\leq\frac{f(\hat{x})-f^*}{\underset{i=1,\ldots,m}{\min}(-g_i(\hat{x}))}$$
From the the information that $\lambda^*$ is optimal solution of (D) we can deduce from that and from farkas non-linear lemma :\ $$f^*\leq q(\lambda^*)=\underset{x}{\min}\{f(x)+\sum_i\lambda_i^*g_i(x)\}$$\ Because we take the minimal $x$ at $q(x^*)$ if we take $\hat{x}$ we only increase the value therefore:\ $\underset{x}{\min}\{f(x)+\sum_i\lambda_i^*g_i(x)\}\leq f(\hat{x})+\sum_i\lambda_i^*g_i(\hat{x})\to f^*\leq f(\hat{x})+\sum_i\lambda_i^*g_i(\hat{x})\iff f(\hat{x})-f^*\geq -\sum_i\lambda_i^*g_i(\hat{x})\geq \underset{i=1,\ldots,m}{\min}(-g_i(\hat{x}))\sum_i\lambda_i^*$\ Therefore we get $f(\hat{x})-f^*\geq\underset{i=1,\ldots,m}{\min}(-g_i(\hat{x}))\sum_i\lambda_i^*$ as desired.\