Consider the convex optimization problem
min $f_0(x)$
s.t $f_i(x)\leq0,i=1,\ldots,m$
where $f_0$ is $C^1$ and convex and $f_i$ are $C^1$ and strictly convex functions.
Let $x^*$ be a feasible solution of the problem.Suppose the following condition is satisfied: there exist $y_i\geq0$,$i\in\{0\}\cup I(x^*)$ which are not all zeroes such that
$y_0\nabla f_0(x^*)+\sum_{i\in I(x^*)}y_i\nabla f_i(x^*)=0$
Prove that $x^*$ is an optimal solution.
$I(x^*)=\{i:f_i(x^*)=0\}$
I've tried to set $S(x)=y_0f_0(x^*)+\sum_{i\in I(x^*)}y_if_i(x^*)$ and then S is convex function as 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 $0=y_0\nabla f_0(x^*)+\sum_{i\in I(x^*)}y_i\nabla f_i(x^*)\iff 0=(y_0\nabla f_0(x^*)+\sum_{i\in I(x^*)}y_i\nabla f_i(x^*))^T(y-x)\iff0=(y_0\nabla f_0(x^*)+\sum_{i\in I(x^*)}y_i\nabla f_i(x^*))^T(y-x)<S(y)-S(x^*)\to S(y)>S(x^*)$ which 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 $y_0f_0(y)+\sum_{i\in I(x^*)}y_if_i(y)\leq y_0f_0(y)$ because $y_i\geq0,f_i(y)\leq0$ and overall we got $y_of_0(y)>y_0f_0(x^*)\iff f_0(y)>f_0(x^*)$ which is the desired result.The strong inequality is because $f_i$ are strictly convex when $y_0\neq0$.
What can I do for the other part?