Finding maximum $b$ in $x^5-20x^4+bx^3+cx^2+dx+e=0$ Let $b, c, d, e$ be real numbers such that the following equation
$$x^5-20x^4+bx^3+cx^2+dx+e=0$$
has real roots only. Find the largest possibe value of $b$.
What I have done is:
Let $x_1, x_2, x_3, x_4, x_5$ be the 5 real roots of the equation. Then we have
$$x_1+x_2+x_3+x_4+x_5=20$$ and
$$b=\sum_{0<i\le j \le 5} x_ix_j=\frac{1}{2}[(x_1+x_2+x_3+x_4+x_5)^2-(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)]$$
To find maximum $b$, we can find minimum $x_1^2+x^2_2+x_3^2+x_4^2+x_5^2$. Cauchy-Schwartz Inequality yields,
$$(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)(1+1+1+1+1)\ge(x_1+x_2+x_3+x_4+x_5)=20$$
Thus,
$$x_1^2+x^2_2+x_3^2+x_4^2+x_5^2\ge\frac{20}{5}=4$$
So,
$$b_{max}=\frac{1}{2}[20^2-4]=198$$
However, the answer was 160, yet I am pretty sure I am correct. Where did I go wrong?
Thanks in advance!

Hint: Let $P(x) = x^5-20x^4+bx^3+cx^2+dx+e$.  If $P$ has all real roots, then $P'''(x)$ must have two real roots..
$\implies 60x^2-480x+6b$ has real roots $\implies b \le 160$
