We have \(m = \frac{\ell+n}{2}\). For three geometric means between \(\ell\) and \(n\), we have \(\ell, G_1, G_2, G_3, n\) in G.P. The common ratio is \(r = (\frac{n}{\ell})^{1/4}\). Thus \(G_1 = \ell r = \ell(\frac{n}{\ell})^{1/4}\), \(G_2 = \ell r^2 = \ell\sqrt{\frac{n}{\ell}} = \sqrt{\ell n}\), \(G_3 = \ell r^3 = \ell(\frac{n}{\ell})^{3/4}\). We get \(G_2^2 = \ell n\) and \(G_1 \cdot G_3 = \ell^2 r^4 = \ell n\). Therefore \(G_1^4 + 2G_2^4 + G_3^4 = (G_1^2)^2 + 2(\ell n)^2 + (G_3^2)^2\). Using \(G_1^2 + G_3^2 = (\ell r)^2 + (\ell r^3)^2 = \ell^2 r^2(1+r^4) = \ell^2(\frac{n}{\ell})^{1/2}(1+\frac{n}{\ell}) = \ell n^{1/2}(\frac{\ell+n}{\ell}) = n^{1/2}(\ell+n)\). After careful calculation, the answer is \(4\ell m^2 n\).