capacitance - Energy dissipated when two charged capacitors are connected in parallel

The question at hand is:

"Two capacitors of capacitances $C_1$ and $C_1$ have charge $Q_1$ and $Q_2$. How much energy, $\Delta w$, is dissipated when they are connected in parallel. Show explicitly that $\Delta w$ is non-negative."

I'm confused about what the physical situation is. I took the assumption that these capacitors were somehow pre-charged, and then connected to each other in parallel without a voltage source. However, I don't understand how this would work. If a circuit is composed of only 2 elements, I don't see how they could be in any arrangement but series. Nevertheless, I tried solving it that way:

Two capacitors in parallel have the same voltage drop. Charge will be redistributed to make it the same voltage for both. Let $Q_1'$ and $Q_2'$ be the charges on the capacitors after they are connected. Now, picture the equivalent capacitor

$C_{eq} = C_1 + C_2 =$ $\frac{Q_1' + Q_2'}{V_f}$

conservation of charge:

$Q_1' + Q_2'= Q_1 + Q_2$,

$C_{eq} =$ $\frac{Q_1 + Q_2}{V_f}$

$V_f = $$\frac{Q_1 + Q_2}{C_1 + C_2}$

The initial energy of the capactiors is:

$U_0 = $$\frac{Q_1^2}{2C_1}$$+\frac{Q_2^2}{2C_2}$

$U_f = $$\frac{1}{2}$$(C_1 + C_2)V_f^2 = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$

$\Delta U$$ = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$-$\frac{Q_1^2}{2C_1}$$-\frac{Q_2^2}{2C_2}$

However, I don't see how this is necessarily non-negative.