Quadratic programming: KKT Optimality conditions I am struggling with an exercice with the following quadratic program:
$$min:x_{1}x_{2} + x^{2}_{1} + \frac{3}{2}x^{2}_{2} + 2x^{2}_{3} + 2x_{1} + x_{2} + 3x_{3}$$
subject to
$$x_{1} + x_{2} + x_{3} = 1$$
$$x_{1} − x_{2} = 0$$
$$x_{1}, x_{2}, x_{3} ≥ 0$$
Is the quadratic objective function convex? Show that $x∗ = (\frac{1}{2}, \frac{1}{2},0)$ is an optimal solution to this problem by ﬁnding vectors $y$ and $s$ that satisfy the optimality conditions jointly with $x∗$.
I already showed that the objective function is convex by proving the positive semidefiniteness of the symmetric matrix $Q$.
My idea for the second part of the question would be to compute the Karush Kuhn Tucker conditions and to check if I place the values for $x*$ in it, that there exists some vectors $y$ and $s$ that fulfill the conditions.
For the Lagrange function I got:
$$min_x:L(x,y,s)=\frac{1}{2}x^TQx+c^Tx-y^T(Ax-b)-s^Tx$$
The first derivative is then:
$$L(x,y,s)'=Qx+c-A^Ty-s$$
How do I proceed with the KKT of the function above?

Let $x_1=x+2=a$. Thus, $c=1-2a$ and
$$x_{1}x_{2} + x^{2}_{1} + \frac{3}{2}x^{2}_{2} + 2x^{2}_{3} + 2x_{1} + x_{2} + 3x_{3}=$$
$$=\frac{23}{2}a^2-11a+5=\frac{23}{2}\left(a^2-\frac{22}{23}a+\frac{10}{23}\right)=$$
$$=\frac{23}{2}\left(a-\frac{11}{23}\right)^2+5-\frac{121}{46}\geq\frac{109}{46}.$$
The equality occurs for $x_1=x_2=\frac{11}{23}$ and $x_3=\frac{1}{23},$ which says that $\frac{109}{46}$ is a minimal value. 
