Use alignat
with manually sized parentheses around the big expression:
\documentclass{article}
\usepackage{amsmath}
\newcommand{\mb}[1]{\mathbf{#1}}
\newcommand{\skp}[2]{\langle#1,#2\rangle}
\newcommand{\cuberoot}[1]{\sqrt[\leftroot{0}\uproot{2}\scriptstyle 3]{#1}}
\begin{document}
\begin{alignat}{3}
&&
-\alpha\skp{\mb{a}}{\mb{n}_\mathcal{W}} & \geq \skp{\mb{a}}{\mb{n}_\mathcal{W}}
\Biggl(&& \cuberoot{\left(\left( 1+ \frac{h_\mathcal{F}}{a_z}\right)^{\!3} -1\right)k_t+\alpha^3}
-\alpha\Biggr) \nonumber \\
&\Leftrightarrow\qquad&
-\alpha &\leq
&&\cuberoot{\left(\left( 1+ \frac{h_\mathcal{F}}{a_z}\right)^{\!3} -1\right)k_t+\alpha^3}
-\alpha \nonumber \\
&\Leftrightarrow&
0 &\leq
&& \left(\left( 1+ \frac{h_\mathcal{F}}{a_z}\right)^{\!3} -1\right)k_t+\alpha^3
\nonumber \\
&\Leftrightarrow&
\alpha &\geq
&&\cuberoot{\left( 1- \left(1+ \frac{h_\mathcal{F}}{a_z}\right)^{\!3} \right)
\frac{\Delta t}{t_\mathcal{F}}} .
\end{alignat}
\end{document}
Personally, I'd not try aligning those terms, leaving them near the inequality they refer to.
Note that your manual spacings \;
are all wrong. There is no need that the outer parentheses fully cover the cube root.
You can make the alignment worse by moving the item in the third line to the right:
\documentclass{article}
\usepackage{amsmath}
\newcommand{\mb}[1]{\mathbf{#1}}
\newcommand{\skp}[2]{\langle#1,#2\rangle}
\newcommand{\cuberoot}[1]{\sqrt[\leftroot{0}\uproot{2}\scriptstyle 3]{#1}}
\newcommand{\cuberootspace}{%
\hphantom{\cuberoot{\vphantom{\left(\left(\frac{h_\mathcal{F}}{a_z}\right)^{3}\right)}}}%
}
\begin{document}
\begin{alignat}{3}
&&
-\alpha\skp{\mb{a}}{\mb{n}_\mathcal{W}} & \geq \skp{\mb{a}}{\mb{n}_\mathcal{W}}
\Biggl(&& \cuberoot{\left(\left( 1+ \frac{h_\mathcal{F}}{a_z}\right)^{\!3} -1\right)k_t+\alpha^3}
-\alpha\Biggr) \nonumber \\
&\Leftrightarrow\qquad&
-\alpha &\leq
&&\cuberoot{\left(\left( 1+ \frac{h_\mathcal{F}}{a_z}\right)^{\!3} -1\right)k_t+\alpha^3}
-\alpha \nonumber \\
&\Leftrightarrow&
0 &\leq
&& \cuberootspace
\left(\left( 1+ \frac{h_\mathcal{F}}{a_z}\right)^{\!3} -1\right)k_t+\alpha^3
\nonumber \\
&\Leftrightarrow&
\alpha &\geq
&&\cuberoot{\left( 1- \left(1+ \frac{h_\mathcal{F}}{a_z}\right)^{\!3} \right)
\frac{\Delta t}{t_\mathcal{F}}} .
\end{alignat}
\end{document}
Or you can improve the appearance by using align
and removing the double arrows
\documentclass{article}
\usepackage{amsmath}
\newcommand{\mb}[1]{\mathbf{#1}}
\newcommand{\skp}[2]{\langle#1,#2\rangle}
\newcommand{\cuberoot}[1]{\sqrt[\uproot{2}\scriptstyle 3]{#1}}
\begin{document}
The following inequalities are easy seen to be equivalent
\begin{align}
-\alpha\skp{\mb{a}}{\mb{n}_\mathcal{W}}
&\geq \skp{\mb{a}}{\mb{n}_\mathcal{W}}
\Biggl(\cuberoot{\biggl(\biggl( 1+ \frac{h_\mathcal{F}}{a_z}\biggr)^{\!3} -1\biggr)k_t+\alpha^3}
-\alpha\Biggr) \nonumber \\
-\alpha &\leq
\cuberoot{\biggl(\biggl( 1+ \frac{h_\mathcal{F}}{a_z}\biggr)^{\!3} -1\biggr)k_t+\alpha^3}
-\alpha \nonumber \\
0 &\leq
\biggl(\biggl( 1+ \frac{h_\mathcal{F}}{a_z}\biggr)^{\!3} -1\biggr)k_t+\alpha^3
\nonumber \\
\alpha &\geq
\cuberoot{\biggl( 1- \biggl(1+ \frac{h_\mathcal{F}}{a_z}\biggr)^{\!3} \biggr)
\frac{\Delta t}{t_\mathcal{F}}} .
\end{align}
\end{document}