Getting square roots right requires a bit of manual intervention.
I essentially add a phantom \beta^K
so TeX will see some more height, but not too much to make it choose the next size of the radical.
Also I add a phantom \Big|
to the denominator, but smashed at the bottom so the apparent height will be increased, thus moving down the radical from the fraction line.
Important suggestion: {^s_i\lambda}
and similar is not the best way to deal with prescripts; with \mathtools
there is \prescript
which will adapt the height of the prescripts to the object they apply to and also adds a thin space in order to make clear that the prescripts aren't scripts to the symbol at the left.
Finally, dcases*
avoids the need for \dfrac
and for \text
.
\documentclass{article}
\usepackage{amsmath,mathtools,amssymb}
\usepackage{bm}
\begin{document}
The original equation is
\begin{equation}
\delta {^s_i\lambda} =
\begin{cases}
\pm \dfrac{{^s\overline{\Delta T}}}{\sqrt[]{\delta {^s_1\bm{u}_l} \cdot \delta {^s_1\bm{u}_l} + \beta}} & \! \text{if} \; i = 1 \\[1em]
- \dfrac{\delta{^s_1\bm{u} \cdot \delta{^s_1\bm{u}_r}}}{\delta {^s_1\bm{u}} \cdot \delta {^s_i\bm{u}_l} + \beta \delta {^s_1\lambda}} & \! \text{otherwise}
\end{cases}
\end{equation}
and now the modified one
\begin{equation}
\delta\prescript{s}{i}{\lambda} =
\begin{dcases*}
\pm \frac{
\prescript{s}{}{\overline{\Delta T}}
}{
\smash[b]{\vphantom{\Big|}}
\sqrt{
\vphantom{\beta^K}
\delta\prescript{s}{1}{\bm{u}_l} \cdot
\delta\prescript{s}{1}{\bm{u}_l} + \beta
}
} & if $i = 1$
\\[1ex]
-\frac{
\delta\prescript{s}{1}{\bm{u}} \cdot \delta\prescript{s}{1}{\bm{u}_r}
}{
\delta\prescript{s}{1}{\bm{u}} \cdot \delta\prescript{s}{i}{\bm{u}_l} +
\beta \delta\prescript{s}{1}{\lambda}
} & otherwise
\end{dcases*}
\end{equation}
\end{document}

\mathstrut
in the square root.