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!

The answer is correct and you are wong.
The mistake in your solution happens when you apply the Cauchy Inequality where you missed to square the right side.
$(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)(1+1+1+1+1)\ge(x_1+x_2+x_3+x_4+x_5)^2=400$
So $x_1^2+x^2_2+x_3^2+x_4^2+x_5^2\ge\frac{400}{5}=80$
and  $b_{max}=\frac{1}{2}[20^2-80]=160$.
A simple check would see your answer is wrong, the third coefficient in $(x-4)^5=160$ and not $198$.
