Consider stationary points of the function $V=px^2+qy^2+rz^2$ subject to the constraints $x^2+y^2+z^2=1$ and $lx+my+nz=0$, where $l,m,n$ not all zero and $p,q,r$ not all equal. How can we show that the stationary values of $V$ satisfy $\frac{l^2}{V-p} + \frac{m^2}{V-q} + \frac{n^2}{V-r} = 0$?
I've shown that the stationary point $(x,y,z)$ must satisfy $l(q-r)yz + m(r-p)xz + n(p-q)xy = 0$, but can't see how to convert this into the required equation. It would be sufficient to show that $(q-r)yz(px^2+qy^2+rz^2-p)=l$ (and similar conditions with $p,q,r$ and $l,m,n$ and $x,y,z$ cycled), but is this true? What about if we add $lx+my+nz$ to the equation I've got?
Thanks for any help with this!