Stumbled upon this one in a textbook:
Let there be a linear transformation $T:V\rightarrow V$ over a finite inner product space $V$. It is known that $TT^* = 7T - 12I$.
How can it be proved that for any $v \in V$ for which $||v||=1$, $||T(v)|| \geq 3$ ?
Note that $TT^{*}$ is self adjoint and so $(7T - 12I) = (7T - 12I)^{*} = 7T^{*} - 12I$ which implies that $T = T^{*}$ so $T$ is also self-adjoint and $T^2 - 7T + 12I = 0$.
Assume that $0 \neq v$ be an eigenvector of $T$ and write $Tv = \lambda v$. Plugging into the equation, we have $(T^2 - 7T + 12I)v = (\lambda^2 - 7\lambda + 12)v = 0$ which implies that $\lambda^2 - 7\lambda + 12 = 0$. Thus, $\lambda = 3$ or $\lambda = 4$.
Finally, since $T$ is self-adjoint, we can take an orthonormal basis $e_1, \ldots, e_n$ of $V$ consisting of eigenvectors of $T$. Write $Te_i = \lambda_i e_i$ where $\lambda_i \in \{ 3, 4 \}$. We then have
$$ \left< Tv, Tv \right> = \left< T\left( \sum_{i=1}^n \left<v, e_i\right> e_i \right), T \left( \sum_{j=1}^n \left< v, e_j \right> e_j \right) \right> = \left< \sum_{i=1}^n \lambda_i \left<v, e_i \right> e_i, \sum_{j=1}^n \lambda_j \left<v, e_j \right> e_j \right> = \sum_{i=1}^n \lambda_i^2 \left|\left< v, e_i \right>\right|^2 \geq \sum_{i=1}^n 9 \left|\left<v, e_i \right>\right|^2 = 9 ||v||^2.$$