Prove the following norm inequality of a vector Prove that $\forall x=(x_1,x_2)\in \mathbb R^2$ satisfying $34x_1^2-24x_1x_2+41x_2^2=25$ we get the following inequality for its norm: $1/\sqrt 2\leq \|x\| \leq1$.
My attempt at solving it: obviously, the given equality references a quadratic form, let's denote it $q$, that is represented by a symmetric matrix that we can easily calculate. Being a real symmetric matrix, it can be diagonalised by some unitary matrix, which most importantly preserves norm. So, we achieve a norm-preserving change of basis matrix that gives us some sort of an equality that is along the lines of $[q(x)]_B=k$ where k is a real number (not sure how to find it) and $B$ is some basis so that $q$ is diagonal in respect to it. This should provide us with enough information needed to prove the inequality. I'm not sure if this is the right approach really. Any help is greatly appreciated.
 A: Your form can be represented as a symmetric matrix:
$$A = \begin{bmatrix}
        34 & -12 \\
        -12 & 41 \\
        \end{bmatrix}$$
$A$ diagonalizes to $\begin{bmatrix}
        50 & 0 \\
        0 & 25 \\
        \end{bmatrix}$ in the basis $\left\{\frac{1}{5}\begin{pmatrix} -3 \\ 4 \end{pmatrix},\frac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix}\right\}$ which happens to be orthonormal.
Introduce the transformation of coordinates:
$$y_1 = \frac{-3x_1+4x_2}{5}$$
$$y_2 = \frac{4x_1+3x_2}{5}$$
Since the basis is orthonormal, the transformation matrix $U = \frac{1}{5}\begin{bmatrix}
        -3 & 4 \\
        4 & 3 \\
        \end{bmatrix}$ is unitary (and symmetric as a bonus), hence the transformation is norm preserving.
The form is reduced to:
$$50y_1^2 + 25y_2^2 = 25$$
$$\frac{y_1^2}{2} + y_2^2 = 1$$
This is an equation of an ellipse centered at the origin with $a = \frac{1}{\sqrt{2}}$ and $b = 1$, so $\|(y_1,y_2)\| \in \left[\frac{1}{\sqrt{2}}, 1\right]$.
Edit:
To diagonalize $A$, first we have to calculate its characteristic polynomial:
$$k_A(\lambda) = \det(A - \lambda I) = \begin{vmatrix}
        34-\lambda & -12 \\
        -12 & 41-\lambda \\
        \end{vmatrix} = (34 - \lambda) (41 - 
\lambda) - 144 = \lambda^2 - 75\lambda + 1250$$
Eigenvalues of $A$ are zeroes of $k_A$:
$$0 = k_A(\lambda) = \lambda^2 - 75\lambda + 1250 \implies \lambda \in \{50,25\}$$
Now we have to find the eigenvectors by solving the homogeneous linear system $(A - \lambda I)x = 0$.
For $\lambda = 25$ we have:
$$\begin{bmatrix}
        9 & -12 \\
        -12 & 16 \\
        \end{bmatrix}\begin{pmatrix} u_1 \\ v_1 \end{pmatrix} = 0$$
The solution is $(u_1,v_1) = \alpha(-3,4)$, for $\alpha \in \mathbb{R}$.
Similarly, for $\lambda = 50$ we have:
$$\begin{bmatrix}
        -16 & -12 \\
        -12 & -9 \\
        \end{bmatrix}\begin{pmatrix} u_2 \\ v_2 \end{pmatrix} = 0$$
The solution is $(u_2,v_2) = \beta(4,3)$, for $\beta \in \mathbb{R}$.
The basis of eigenvectors we got is $\left\{\begin{pmatrix} -3 \\ 4 \end{pmatrix},\begin{pmatrix} 4 \\ 3 \end{pmatrix}\right\}$. We have to orthonormalize it. Since the eigenvectors are already orthogonal (because the matrix was Hermitian/symmetric), all we have to do is normalize them: that is how we get the orthonormal basis $\left\{\frac{1}{5}\begin{pmatrix} -3 \\ 4 \end{pmatrix},\frac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix}\right\}$. In general though, we would have to use the Gram-Schmidt orthonormalization process on every eigenspace (the eigenspaces are orthogonal to each other).
We set the eigenvectors from the basis as columns of our (unitary) transformation matrix $U = \frac{1}{5}\begin{bmatrix}
        -3 & 4 \\
        4 & 3 \\
        \end{bmatrix}$.
Now we obtain the diagonalization as:
$$\begin{bmatrix}
        50 & 0 \\
        0 & 25 \\
        \end{bmatrix} = U^{-1}AU = U^*AU = UAU$$
Now just express a vector $(x_1,x_2)$ in this basis to get the new coordinates $y_1, y_2$:
$$\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = U\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$
Since $U$ is also symmetric, as a bonus we also get:
$$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = U^*\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}=U\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$$
A: Let $x_1=r\cos\theta,x_2=r\sin\theta$. Then one has
$$ r^2(34\cos^2\theta-12\sin(2\theta)+41\sin^2\theta)=25 $$
which gives
$$ r^2(34\cdot\frac12(1+\cos(2\theta))-12\sin(2\theta)+41\cdot\frac12(1-\sin(2\theta))=25. $$
Thus
$$ r^2=\frac{50}{75-7\cos(2\theta)-24\sin(2\theta)}. $$
Note that
$$ |-7\cos(2\theta)-24\sin(2\theta)|\le\sqrt{7^2+24^2}=25 $$ 
and hence
$$ \frac12 \le r^2\le 1 $$
or 
$$ \frac{1}{\sqrt2}\le\|x\|\le 1. $$
