The target expression resembles quadratic root. So let $x_{1,2}=c\pm\sqrt{c^2-ab}$. Then $x_1$ and $x_2$ are the two roots of $$x^2-2cx+ab=0$$ \begin{align*} \therefore\quad (x_1-a)(x_2-a) &= x_1x_2-(x_1+x_2)a + a^2\\ &= ab -2ca + a^2\\ &= a(b+a-2c)\\ &< 0&\scriptsize{(\because\quad a>0,\ 2c>a+b)} \end{align*} This means that $(x_1-a)$ and $(x_2-a)$ have different signs. Clearly, $x_1 > x_2$. Hence, we have $x_2 < a < x_1$ which means $$c-\sqrt{c^2-ab} < a < c+\sqrt{c^2-ab}$$