$$\begin{align} v(p_1, p_2, w) & = \sqrt{\frac w{p_1^2\left(\frac1{p_1}+\frac1{p_2}\right)}} + \sqrt{\frac w{p_2^2\left(\frac1{p_1}+\frac1{p_2}\right)}} \\ & = \sqrt{\frac w{\left(\frac1{p_1}+\frac1{p_2}\right)}} \left(\sqrt{\frac1{p_1^2}}+\sqrt{\frac1{p_2^2}}\right) \\ & = \sqrt{\frac w{p_1}+\frac w{p_2}} \end{align}$$
So as the title says, can anyone explain to me this square root?
Step by step:
Rewrite the original expression as follows:
$$\sqrt{\frac{w}{p_{1}^{2}(\frac{1}{p_{1}} + \frac{1}{p_{2}})}} + \sqrt{\frac{w}{p_{2}^{2}(\frac{1}{p_{1}} + \frac{1}{p_{2}})}} = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}}\right) + \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{2}^{2}}}\right)$$
Factoring, this becomes
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}}\right) + \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{2}^{2}}}\right) = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}} + \sqrt{\frac{1}{p_{2}^{2}}}\right)$$
Note that
$$\sqrt{\frac{1}{p_{1}^{2}}} = \frac{1}{p_{1}}$$
and
$$\sqrt{\frac{1}{p_{2}^{2}}} = \frac{1}{p_{2}}$$
So
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}} + \sqrt{\frac{1}{p_{2}^{2}}}\right) = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)$$
This can be rewritten as
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right) = \sqrt{\frac{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)^{2}}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}$$
And finally, we can simplify this to
$$\sqrt{\frac{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)^{2}}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}} = \sqrt{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)} = \sqrt{\frac{w}{p_{1}} + \frac{w}{p_{2}}}$$
Substituting $q_i=1/p_i$ we find $$q_1\sqrt{\frac{w}{q_1+q_2}}+q_2\sqrt{\frac{w}{q_1+q_2}}=(q_1+q_2)\sqrt{\frac{w}{q_1+q_2}}=\sqrt{w(q_1+q_2)},$$ as $q_1+q_2=\sqrt{(q_1+q_2)^2}$.
