If I have the following equation:
$$f = 2\gamma_2 \sigma_{2,1}^2 a^2 + 4 \gamma_2\sigma_{2,1}\sigma_{2,2}ax_2 + 2\gamma_2 \sigma_{2,2}^2x_2^2$$
I can write it as:
$$f = 2\gamma_2(\sigma_{2,1}a + \sigma_{2,2}x_2)^2$$
But now I have:
$$g = 2\gamma_1 \sigma_{2,1}^2 a^2 + 2 (\gamma_1 + \gamma_2)\sigma_{2,1}\sigma_{2,2}ax_2 + 2\gamma_2 \sigma_{2,2}^2x_2^2$$
Now I'd like to complete the squares, but I don't see how.
\begin{align} g &= 2\gamma_1 \sigma_{2,1}^2 a^2 + 2 (\gamma_1 + \gamma_2)\sigma_{2,1}\sigma_{2,2}ax_2 + 2\gamma_2 \sigma_{2,2}^2x_2^2 \\ &= 2(\gamma_1 \sigma_{2,1} a + \gamma_2 \sigma_{2,2} x_2) ( \sigma_{2,1} a + \sigma_{2,2} x_2) \end{align}
Things are occurring in groups. Acknowledging them might clear away some of the noise.
Let $A = \sigma_{2,1} a$ and let $B = \sigma_{2,2} x_2$.
Then $g = 2(\gamma_1 A + \gamma_2 B)(A + B)$
You said you wanted to prove, if possible, that $\gamma_1 \ge 0$ and $\gamma_2 \ge 0$ implies $g \ge 0$. I'm going to have to assume that $A$ and $B$ can be any real number.
But if $\gamma_1 > 0$ and $\gamma_2 = 0$, then you get $g = 2 \gamma_1 A(A + B)$ which you can make negative if you make $A$ positive and $A+B$ negative.
So it seems that your hypothesis is false; unless there are things about $\sigma_{2,1} a$ and $\sigma_{2,2} x_2$ that you haven't told us about.