Problem: Let $T$ be a positive definite selfadjoint operator in an $n$-dimensional inner product space $H$. Find the maximum possible angle between a vector $v\neq 0$ and its image $Tv$, expressed in terms of the eigenvalues $\theta_0^2\geq\dots\geq\theta_{n-1}^2>0$ of $T$.
Solution: We want to minimize $\cos(v,Tv)=\frac{\langle v,Tv \rangle}{\|v\|\|Tv\|}$, or equivalently, its square $\frac{\langle v,Tv\rangle^2}{\langle v,v\rangle\langle Tv,Tv\rangle}$.
Since the angle between any vector $v\neq 0$ and its image $Tv$ doesn't change if we rescale $v$, we can rescale at our wish. I choose to restrict to those $v$ such that the numerator $g(v)=\langle v,Tv\rangle$ equals 1, so now the problem is equivalent to minimizing the denominator $f(v)=\langle v,v\rangle\langle Tv,Tv\rangle$, and now there are no divisions bothering.
Critical points of this restricted function are given by the equation $df(v)=\lambda dg(v)$, where $\lambda\in\mathbb R$ is a Lagrange multiplier. We calculate
$dg(v)=\langle v,T-\rangle+\langle -,Tv\rangle=2\langle Tv,-\rangle$ and
$df(v)=2\langle v,-\rangle\langle Tv,v\rangle+2\langle v,v\rangle\langle Tv,T-\rangle=2\|Tv\|^2\langle v,-\rangle+2\|v\|^2\langle T^2v,-\rangle$.
The critical point equation can now be rewritten:
$\langle \|Tv\|^2v+\|v\|Tv-\lambda Tv,-\rangle=0$
so $v$, $Tv$, and $T^2v$ are linearly dependent when $v$ is a critical point. But we know by Vandermonde that they would be independent if $v$ had nonzero projections in three eigenspaces[...]. So we can restrict to each plane spanned by two eigenvectors, where we must solve our problem in the case $n=2$, being only interesting the case in which the two eigenvalues are different, because otherwise $v$ is also an eigenvector and the angle will be zero. Now I have little time to write, but the problem can be solved by single variables methods (parametrizing the circle with the usual $\cos$ and $\sin$).
Calculations seem to get a little nicer if we express $T=S^2$, by letting $S$ be the only positive selfadjoint square root of $T$, with eigenvalues $\theta_0\geq\dots\theta_{n-1}>0$. We then have $g(v)=\langle Sv,Sv\rangle=\|Sv\|^2$, and $f(v)=\|v\|^2\|S^2v\|^2$.
Also, the symmetry is greater if we work in terms in terms of the variable $w=Sv$, so that $g=\|w\|^2$ and $f=\|S^{-1}w\|^2\|Sw\|^2$. If $w=\alpha v_i+ \beta v_j$ were $v_i$ and $v_j$ are eigenvectors of $S$ with eigenvalues $\theta_i$ and $\theta_j$ and $i\leq j$, the four critical points can be found after some calculations, and they are given by $\alpha^2=\beta^2=\frac 12$. The cosine of the angle is then $\frac{\langle w,w\rangle}{\|S^-1w\|\|Sw\|}=\frac{\theta_i\theta_j}{2(\theta_i^2\theta_j^2)}=\frac 12 (\frac{\theta_i}{\theta_j}+\frac{\theta_j}{\theta_i})$.
Once solved the case $n=2$, we want to select $i$ and $j$ so that $(\frac{\theta_i}{\theta_j}+\frac{\theta_j}{\theta_i})$ is maximum. But the function $h(t)=t+t^{-1}$ increases in the interval $[1,+\infty)$ so the value $t=\frac{\theta_i}{\theta_j}$ should be chosen as large as possible by maximizing $\theta_i$ and minimizing $\theta_j$.