Finding maximum $b$ in $x^5-20x^4+bx^3+cx^2+dx+e=0$

Question

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!

Answer 1

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 $198$.

Answer 2

The condition for your answer is when $x_1=x_2=...=x_5=4$,so we can find that they are roots of $$(x-4)^5=0$$

Expand this equation of left side we have the coefficient of $x^4$ is $$C(5,4)*(-4)^1=-20$$ and coefficient of $x^3$ is $$C(5,3)*(-4)^2=160$$

Answer 3

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 $\implies b \le 160$