Let $a, b$ be real numbers such that all roots of $f(x) := x^4 – x^3 + ax + b$ are real. Prove that $f(-1/2) \le 3/16$.
The question was posted recently which was closed,
due to missing of contexts etc.
My attempt: $f(-1/2) \le 3/16$ is equivalent to $a \ge 2b$.
Let $x_1, x_2, x_3, x_4$ be the real roots of $f(x)$.
By Vieta, we have
\begin{align*}
x_1 + x_2 + x_3 + x_4 &= 1, \\
x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 &= 0, \\
x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 &= -a, \\
x_1x_2x_3x_4 &= b.
\end{align*}
The problem becomes:
If $x_1, x_2, x_3, x_4$ are real numbers such that
$x_1 + x_2 + x_3 + x_4 = 1$ and $x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 = 0$,
prove that
$$x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 + 2x_1x_2x_3x_4 \le 0.$$
This is true (e.g. verified by Mathematica). Is there a nice proof for it (or the original problem)?
Best Answer
Suppose $b= a/2+c$ where $c>0$. Then we have $$\underbrace{x^4-x^3+ax+{a\over 2}}_{p(x)} =-c$$ This means that graph of $p$ and line $y=-c<0$ have at least two different common points (if it has only one then $p(x)=(x-d)^4-c$ for some real $d$, this case is not possible) so $p$ has exactly two minimums with negative value, say at $x_1$ and $x_3$ where $x_1<x_3$, so $p(x_1)<0$ and $p(x_3)<0$ i.e. $$x_i^4-x_i^3+ax_i+{a\over 2} <0\;\;\;(*)$$ for $i=1,3$. Since $x_1,x_3$ satisfies also $p'(x_i)=0$ we have $$a=-4x_i^3+3x_i^2\;\;\;(**)$$ so we get plugging $(**)$ in inequality $(*)$ for $i=1,3$: $$-6x_i^4+3x_i^2<0\implies x_i^2>{1\over 2}$$
If we subtract equations $(**)$, we get $$(x_1-x_3)\Big(4(x_1^2+x_1x_3+x_3^2)-3(x_1+x_3)\Big)=0$$ so $$4(x_1^2+x_1x_3+x_3^2)=3(x_1+x_3)$$ and from here we get $$2(x_1^2+x_3^2)+2(x_1+x_3)^2=3(x_1+x_3)$$ which means that $$ 2+2t^2<3t$$ where $t=x_1+x_3$. But this inequality does not have a solution so we have a contradiction.