Elementary matrix/polynomial problem involving inequalities I expect this probably has an easy answer but I'm feeling stupid and I'm stumped.
A book I'm reading, if I've understood it correctly, includes the following assertion. Let $M$ be a $3\times3$ symmetric matrix of real numbers with the following form:
$M \equiv \left(\begin{array} \ 1 \ a \ b \\ a \ 1 \ c \\ b \ c \ 1 \end{array}\right)$
Then $M$ is positive semidefinite iff $a^2 + b^2 + c^2 - 2 a b c \leq 1$.
The only approach I can think of is to look at the characteristic equation of $M$, which (if I didn't mess up) turns out to be
$(1 - \lambda)^3 - (a^2 + b^2 + c^2)(1 - \lambda) + 2 a b c$
So it seems that I need to show that, if $\alpha >= 0$ and $\beta \in \mathbb{R}$, then $\alpha - \beta \leq 1$ iff all roots of $x^3 - \alpha x + \beta$ satisfy $x \leq 1$. But I can't see how to prove that. Can anyone help?

$A$ is positive semidefinite $\iff$ all eigenvalues of $A$ are
greater than $0$. To show semidefiniteness, we look at the bilinear
form 
$x^{T}Ax$
If this quantity is $\geq0$, then $A$ is positive semidefinite.
Note that we can use the Spectral theorem which states that every
$n\times n$ symmetric matrix has $n$ linearly independent eigenvectors.
This means that we can use a change of basis to rewrite $A$ in terms
of its eigenvalues and eigenvectors (or equivalently, we can rewrite
any matrix as $Q\Lambda Q^{-1}=Q\Lambda Q^{T}$ for symmetric matrices
with eigenvectors chosen to be orthogonal). We can equivalently rewrite
any vector $x\in\mathbb{R}^{n}$ in terms of these eigenvectors, upon
which 
$
x^{T}Ax=x^{T}\lambda x=\lambda\left\langle x,x\right\rangle \geq0\ \iff\lambda\geq0
$
Since the inner product of two vectors is a positive definite bilinear
form, we only require that $\lambda\geq0$. Without loss of generality,
we can use linearity to only have to show this for each eigenvalue. 
But note that $\det A=\lambda_{1}\cdots\lambda_{n}$. However, we
note that for the case of $n=3$, $\det A\geq0$ implies that either
all of the $\lambda_{j}>0$ or $2$ of the $\lambda_{j}&lt0$. You just
need one more step showing that this latter case does not happen,
or something equivalent to show that $\det A\geq0\implies\lambda_{j}>0$.
Once you have done this, you can see that all you need to show is
that $\det A\geq0$, which would imply that $1-(a^{2}+b^{2}+c^{2})+2abc\geq0$,
which is equivalent to the assertion. Hope this helps!
