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 $b = x_1 x_2 +  x_1 x_3 + \dots + x_4 x_5$.
 A: We will prove the slightly stronger statement

Claim: Let
  $$
p(x) = x^n + ax^{n-1} + bx^{n-2} + \cdots + c
$$
  be a polynomial with real coefficients. If $n=2,3,4,5$ and $2a^2 < 5b$ then $p$ has at least one non-real zero.
If $n \geq 6$ then $p$ may have all real zeros regardless of whether $2a^2 < 5b$ or $2a^2 \geq 5b$. For example, the polynomial
  $$
\left(x - \sqrt{\frac{2}{n(n-1)}}\right)^n = x^n - \sqrt{\frac{2n}{n-1}} x^{n-1} + x^{n-2} + \cdots + \left(\frac{2}{n(n-1)}\right)^n
$$
  has all real zeros with $2a^2 < 5b$.

(In fact $p$ has at least two real zeros since they will come in complex conjugate pairs.)
As noted in the question, if $x_1,\ldots,x_n$ are the zeros of $p$ then
$$
a = -\sum_{k=1}^{n}x_k \qquad \text{and} \qquad b = \sum_{1 \leq j < k \leq n} x_j x_k,
$$
so that
$$
\begin{align}
2a^2 &= 2\sum_{k=1}^{n} x_k^2 + 4\sum_{1 \leq j < k \leq n} x_j x_k \\
&= 2\sum_{k=1}^{n} x_k^2 + 4b
\end{align}
$$
or, rearranging,
$$
2a^2 - 5b = 2\sum_{k=1}^{n} x_k^2 - \sum_{1 \leq j < k \leq n} x_j x_k.
$$
To prove your statement it therefore suffices to show that
$$
2\sum_{k=1}^{n} x_k^2 \geq \sum_{1 \leq j < k \leq n} x_j x_k \tag{1}
$$
for $x_1,\ldots,x_n \in \mathbb R$ and $n = 5$.
It's clearly true if
$$
\sum_{1 \leq j < k \leq n} x_j x_k \leq 0,
$$
so suppose that
$$
\sum_{1 \leq j < k \leq n} x_j x_k > 0
$$
and set
$$
y_\ell = \left( \sum_{1 \leq j < k \leq n} x_j x_k \right)^{-1/2} x_\ell, \qquad \ell = 1,\ldots,n.
$$
Then
$$
\sum_{k=1}^{n} x_k^2 = \left(\sum_{1 \leq j < k \leq n} x_j x_k \right) \sum_{k=1}^{n} y_k^2,
$$
so that $(1)$ becomes
$$
2 \sum_{k=1}^{n} y_k^2 \geq 1. \tag{2}
$$
We also note that
$$
\sum_{1 \leq j < k \leq n} y_j y_k = 1.
$$
Our mode of attack to prove the new inequality $(2)$ will be to use Lagrange multipliers to minimize $\sum_{k=1}^{n} y_k^2$ subject to this constraint. To this end, define
$$
f(y_1,\ldots,y_n,\lambda) = \sum_{k=1}^{n} y_k^2 + \lambda\left(\sum_{1 \leq j < k \leq n} y_j y_k - 1\right).
$$
We calculate
$$
f_{y_k}(y_1,\ldots,y_n,\lambda) = 2y_k + \lambda \sum_{j\neq k} y_j = (2-\lambda) y_k + \lambda \sum_{j=1}^{n} y_j, \qquad k=1,\ldots,n,
$$
and thus we need to solve the $n+1$ equations
$$
\begin{align}
0 &= (2-\lambda) y_k + \lambda \sum_{j=1}^{n} y_j, \qquad k=1,\ldots,n, \\
1 &= \sum_{1 \leq j < k \leq n} y_j y_k.
\end{align}
$$
If $\lambda \neq 2$ then
$$
y_k = \frac{\lambda}{\lambda-2} \sum_{j=1}^{n} y_j
$$
and so $y_1 = y_2 = \cdots = y_n$, and if $y_k \neq 0$ then this yields
$$
\lambda = \frac{2}{1-n}.
$$
Further, the equation $1 = \sum_{1 \leq j < k \leq n} y_j y_k$ becomes
$$
y_k = \sqrt{\frac{2}{n(n-1)}}.
$$
We will omit checking that this is a local minimum of the problem. Finally, toward $(2)$ we calculate
$$
2\sum_{k=1}^{n} y_k^2 = \frac{4}{n-1} \geq 1
$$
for $n = 2,3,4,5$, from which the claim, and the result in the question, follows. In particular, taking $x_\ell = y_\ell$ furnishes the counterexample at the end of the claim.
A: 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 $2a^2 \geq 5b$.
