Let $x^*$ be a feasible solution of the following convex optimization problem \begin{align}\min&\quad f_0(x)\\\text{s.t.}&\quad f_i(x)\leq0,i=1,\ldots,m\end{align} where $f_0$ is $C^1$ and convex and $f_i$ are $C^1$ and strictly convex functions. Suppose the following condition is satisfied:
- there exist $y_i\geq0$ for $i\in\{0\}\cup I(x^*)$ where $I(x^*)=\{i:f_i(x^*)=0\}$ which are not all zeroes such that $y_0\nabla f_0(x^*)+\sum\limits_{i\in I(x^*)}y_i\nabla f_i(x^*)=0$.
Prove that $x^*$ is an optimal solution.
I've tried to set $S(x)=y_0f_0(x^*)+\sum\limits_{i\in I(x^*)}y_if_i(x^*)$. Then $S$ is a convex function as it is a sum of convex functions and $x^*$ is minimizer of $S$ because $\nabla S(x^*)=0$ and $$f(x^*)=y_0f_0(x^*)+\sum_{i\in I(x^*)}y_if_i(x^*)=S(x^*).$$ Using the gradient inequality we get $0=y_0\nabla f_0(x^*)+\sum_{i\in I(x^*)}y_i\nabla f_i(x^*)$ if and only if $$0=\left(y_0\nabla f_0(x^*)+\sum_{i\in I(x^*)}y_i\nabla f_i(x^*)\right)^\top(y-x)<S(y)-S(x^*)$$ which implies that $S(y)>S(x^*)$. This is the same as $$y_0f_0(y)+\sum_{i\in I(x^*)}y_if_i(y)>y_0f_0(x^*)+\sum_{i\in I(x^*)}y_if_i(x^*)=y_0f_0(x^*)$$ and $y_0f_0(y)+\sum_\limits{i\in I(x^*)}y_if_i(y)\leq y_0f_0(y)$ because $y_i\geq0$ and $f_i(y)\leq0$. Thus $$y_0f_0(y)>y_0f_0(x^*)\iff f_0(y)>f_0(x^*)$$ which is the desired result. The strong inequality is because $f_i$ are strictly convex.
Where $y_0\neq0$
Now I need to solve the case when $y_0=0$ any hint?