Finding conditions on the eigenvalues of a matrix Consider the $2\times2$ matrix 
$$A = \begin{pmatrix}a&b\\c&d\end{pmatrix}$$
where $a,b,c,d\ge 0$. Show that $\lambda_1\ge\max(a,d)>0$ and $\lambda_2\le\min(a,d)$.

So the eigenvalues are given by the characteristic polynomial
$$(a-\lambda)(d-\lambda)-bc=0\implies \lambda^2 - (a+d)\lambda + ad - bc=0$$
And so the solutions to this equation are
$$\lambda_{1,2} = \frac{(a+d)\pm\sqrt{(a+d)^2-4(ad-bc)}}2 = \frac{(a+d)\pm\sqrt{(a-d)^2+4bc}}2$$
We may therefore simplify this to:
$$\lambda_1+\lambda_2=a+d$$
But now how would one simplify this into the conditions above?

Note that by the formula you derived one of the conditions implies the other. So suppose $\lambda_1\geq \lambda_2$. Now you know that
$$ \begin{pmatrix}
a-\lambda_1 & b\\
c & d-\lambda 1
\end{pmatrix} $$ 
is singular, i.e. there is a real number $r$ s.t. 
$$ \begin{pmatrix}
a-\lambda_1 \\
c
\end{pmatrix} =r\begin{pmatrix}
 b\\
 d-\lambda_1
\end{pmatrix} . $$ 
the case $a=d$ is easy because then $2\lambda_1\geq \lambda_1+\lambda_2=2a$ which implies your claim.
Now suppose that $a > d$ and $a>\lambda_1\geq d$. It is enough to consider this case, since $a>d\geq \lambda_1$ would imply that $\lambda_2\geq a>d$.
Now you know that $a-\lambda_1=rb$ and therefore $b\neq 0$. So you conclude that $r=(a-\lambda_1)/b$. But now this is a positive number and in order to have $c=r(d-\lambda_1)$ you must conclude that $c=d-\lambda_1=0$ and therefore $d=\lambda_1$. But this contradicts $\lambda_1 \geq \lambda_2$.
