proving the existence of a complex root If 
$x^5+ax^4+ bx^3+cx^2+dx+e$ where $a,b,c,d,e \in {\bf R}$ 
and 
$2a^2< 5b$
then the polynomial has at least one non-real root.
We have $-a = x_1 + \dots + x_5$ and $b = x_1 x_2 +  x_1 x_3 + \dots + x_4 x_5$.

Here is a simple proof.
Let's prove this by contraposition: if $x_1, x_2, x_3, x_4, x_5$ are all real, then $2a^2 \geq 5b$ should hold.
You have a formula for $-a$ and $b$ already; from these you can easily calculate that
$$2a^2-5b = \left( 2x_1^2 + 2x_2^2 + 2x_3^2 + 2x_4^2 + 2x_5^2 \right) - x_1x_2 - x_1x_3 - \dots - x_4x_5.$$
Double it once more, and it starts to remind us of something:
$$2(2a^2-5b) = \color{tomato}{\left( 4x_1^2 + 4x_2^2 + 4x_3^2 + 4x_4^2 + 4x_5^2 \right)} \color{steelblue}{- 2x_1x_2 - 2x_1x_3 - \dots - 2x_4x_5}.$$
We have a bunch of $x_i^2 + x_j^2$ and a bunch of $-2x_ix_j$, which looks like the expansion of $(x_i - x_j)^2$, doesn't it? Using this clever insight, we can rewrite this expression into a sum of squares. For each term subtracted in the blue half, we combine it the squares it needs from the red half. Since each $x_i \in \{x_1, x_2, x_3, x_4, x_5\}$ occurs exactly four times in the blue half, all of our squares in the red half nicely vanish. All we are left with is
\begin{align*}
2(2a^2-5b) &= \left(\color{tomato}{x_1^2} \color{steelblue}{-2x_1 x_2} \color{tomato}{+x_2^2}\right) + \left(\color{tomato}{x_1^2} \color{steelblue}{-2x_1 x_3} \color{tomato}{+x_3^2}\right) + \dots
\\ &= (x_1-x_2)^2 + (x_1-x_3)^2 + \dots + (x_4-x_5)^2 
\end{align*}
(If you're not convinced, try expanding this back.)
Since we've written $2(2a^2-5b)$ as a sum of squares, clearly it is $\geq 0$. We can conclude that $2a^2 \geq 5b$.
