Proving an inequality in Eigenvalues of a matrix If $\lambda_1$, $\lambda_2$, $\lambda_3$ are the eigenvalues of the matrix : $$
        \begin{pmatrix}
        26 & -2 & 2 \\
        2 & 21 & 4 \\
        4 & 2 & 28 \\
        \end{pmatrix}
$$ Show that : $$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} \le \sqrt{1949}$$ 
I found the characterstic equation as $$\lambda^3 - 75\lambda^2 + 1850\lambda - 15000 = 0$$
This gave $$\lambda_1 + \lambda_2 + \lambda_3 = 75$$ and $$\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_1\lambda_3 = 1850$$
Also $${(\lambda_1 + \lambda_2 + \lambda_3)}^2 = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 + 2{(\lambda_1\lambda_2 + \lambda_2\lambda_3 + \lambda_1\lambda_3)}$$
From this I found out $$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} = \sqrt{1925}$$
But am not satisfied with the way I have approached the problem. Does anyone know of an easier way of doing it?
 A: OK, another one, to give a hint where the $1949$ comes from:
If your Matrix is $A=((a_{ij}))_{i,j=1,\ldots,n}$ with real entries, then the Frobenius norm of $A$ is defined as
$$\|A\|_{\mathrm{F}} = \sqrt{\operatorname{tr}(A^{\mathsf{T}} A)} =
\sqrt{\sum_{i,j=1,\ldots,n} a_{ij}^2}$$
Note that
$$\sum_{i,j=1,\ldots,n} a_{ij}^2 - \sum_{i,j=1,\ldots,n} a_{ij}\,a_{ji} =
\sum_{i,j=1,\ldots,n;i<j} (a_{ij} - a_{ji})^2\geq 0$$
In other words,
$$\operatorname{tr}(A^2) \leq \operatorname{tr}(A^{\mathsf{T}} A)$$
which in your case means
$$\sum_{i=1}^n\lambda_i^2 \leq 1949$$
So this is where the $1949$ comes from, it's the square of the Frobenius norm of $A$.
Edit: typo corrections
A: Hint: If your matrix is $A$ with eigenvalues $\lambda_1,\ldots,\lambda_n$, then
$$\operatorname{tr} A^2 = \sum_{i=1}^n \lambda_i^2$$
A: The question seems to me just ask a bound and not an exact value that's we  can found by a little calculus. However there's a method which localize the eigenvalues of a squared matrix and called Gershgorin circle theorem. Let's apply this method in our example: we obtain the three discs:
$$D(26,4)\quad;\quad D(21,4)\quad;\quad D(28,6)$$
and then by triangle inequality we find
$$\sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2} \le \sqrt{30^2+25^2+34^2}=\sqrt{2681}$$ 
